NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

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SHORT ANSWER TYPE QUESTIONS

Q.1. Find the equation of the circle which touches the both axes in first quadrant and whose radius is a.
Ans.
Clearly centre of the circle = (a, a)
and radius = a
Equation of circle with radius r and centre (h, k) is
(x – h)2 + (y – k)2 = r2
So, the equation of the required circle
⇒ (x – a)2 + (y – a)2 = a2
⇒ x2 – 2ax + a2 + y2 – 2ay + a2 = a2
⇒ x2 + y2 – 2ax – 2ay + a2 = 0
Hence, the required equation is
x2 + y2 – 2ax – 2ay + a2 = 0
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

Q.2. Show that the point (x, y) given by x =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev lies on a circle for all real values of t such that –1 ≤  t ≤ 1 where a is any given real numbers.
Ans.
GivenNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
= a2
∴ x2 + y2 = a2 which is the equation of a circle.
Hence, the given points lie on a circle.

Q.3. If a circle passes through the point (0, 0) (a, 0), (0, b) then find the coordinates of its centre.
Ans.
Given points are (0, 0), (a, 0) and (0, b)
General equation of the circle is x2 + y2 + 2gx + 2fy + c = 0
where the centre is (– g, – f) and radius =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
If it passes through (0, 0)
∴ c = 0
If it passes through (a, 0) and (0, b) then
a2 + 2ga + c = 0 ⇒ a2 + 2ga = 0 [∵ c = 0]
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
and 0 + b2 + 0 + 2fb + c = 0 ⇒ b2 + 2fb = 0 [∵ c = 0]
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the coordinates of centre of the circle are (– g, – f)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

Q.4. Find the equation of the circle which touches x-axis and whose centre is (1, 2).
Ans. Since the circle whose centre is (1, 2) touch x-axis
∴ r = 2
So, the equation of the circle is
(x – h)2 + (y – k)2 = r2
   (x – 1)2 + (y – 2)2 = (2)2
⇒ x2 – 2x + 1 + y2 – 4y + 4 = 4
  x2 + y2 – 2x – 4y + 1 = 0
Hence, the required equation is
x2 + y2 – 2x – 4y + 1 = 0.
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

Q.5. If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle. 

[Hint: Distance between given parallel lines gives the diameter of the circle.]
Ans. Given equation are 3x – 4y + 4 = 0
and 6x – 8y – 7 = 0 ⇒NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
SinceNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRevthen the lines are parallel.
So, the distance between the parallel lines

NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Diameter = 3/2
∴ Radius = 3/4
Hence, the required radius = 3/4.

Q.6. Find the equation of a circle which touches both the axes and the line 3x – 4y + 8 = 0 and lies in the third quadrant. 
[Hint: Let a be the radius of the circle, then (– a, – a) will be centre and perpendicular distance from the centre to the given line gives the radius of the circle.]
Ans. Let a be the radius of the circle.
Centre of the circle = (– a, – a)
Distance of the line 3x – 4y + 8 = 0
From the centre = Radius of the circle
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ a = 5a – 8   
⇒ 5a – a = 8
⇒ 4a = 8 ⇒ a = 2
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
and
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
∴ The equation of the circle is
(x + 2)2 + (y + 2)2 = (2)2
  x2 + 4x + 4 + y2 + 4y + 4 = 4
  x2 + y2 + 4x + 4y + 4 = 0
Hence, the required equation of the circle
x2 + y2 + 4x + 4y + 4 = 0.

Q.7. If one end of a diameter of the circle x2 + y2 – 4x – 6y + 11 = 0 is (3, 4), then find the coordinate of the other end of the diameter.
Ans. Let the other end of the diameter is (x1, y1).
Equation of given circle is
x2 + y2 – 4x – 6y + 11 = 0
Centre = (– g, – f) = (2, 3)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the required coordinates are (1, 2).

Q.8. Find the equation of the circle having (1, –2) as its centre and passing through 3x + y = 14, 2x + 5y = 18
Ans. Given equations are
3x + y = 14 ...(i)
and 2x + 5y = 18 ...(ii)
From eq. (i) we get y = 14 – 3x ...(iii)
Putting the value of y in eq. (ii) we get
⇒ 2x + 5(14 – 3x) = 18    
⇒ 2x + 70 – 15x = 18   
⇒ – 13x = – 70 + 18
⇒ – 13x = – 52
∴ x = 4
From eq. (iii) we get,
y = 14 – 3 x 4 = 2
∴ Point of intersection is (4, 2)
Now, radius r =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
So, the equation of circle is (x – h)2 + (y – k)2 = r2
  (x – 1)2 + ( + 2)2 = (5)2
⇒ x2 – 2x + 1 + y2 + 4y + 4 = 25
⇒ x2 + y2 – 2x + 4y – 20 = 0
Hence, the required equation is x2 + y2 – 2x + 4y – 20 = 0

Q.9. If the line y = √3x + k touches the circle x2 + y2 = 16, then find the value of k. 
[Hint: Equate perpendicular distance from the centre of the circle to its radius].
Ans. Given circle is x2 + y2 = 16
Centre = (0, 0)
radius r = 4
Perpendicular from the origin to the given line y = √3x + k is equal to the radius.
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the required values of k are ± 8.

Q.10. Find the equation of a circle concentric with the circle x2 + y2 – 6x + 12y + 15 = 0 and has double of its area. 
[Hint: concentric circles have the same centre.]
Ans. Given equation of the circle is
x2 + y2 – 6x + 12y + 15 = 0     ...(i)
Centre = (– g, – f) = (3, – 6)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Since the circle is concentric with the given circle
∴ Centre = (3, – 6)
Now let the radius of the circle is r
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Area of the given circle (i) = πr2 = 30π sq unit
Area of the required circle = 2 x 30π = 60π sq. unit
If r1 be the radius of the required circle 2
πr12 = 60π ⇒ r21 = 60
So, the required equations of the circle is
(x - 3)2 + (y + 6)2 = 60
 x2 + 9 - 6x + y2 + 36 + 12y - 60 = 0
⇒ x2 + y2 – 6x + 12y – 15 = 0
Hence, the required equation is x2 + y2 – 6x + 12y – 15 = 0.

Q.11. If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.
Ans. Let the equation of an ellipse is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Length of major axis = 2a
Length of minor axis = 2b.
and the length of latus rectum =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
According to the question, we have
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Now b2 = a2(1 – e2),
where e is the eccentricity
⇒ b2 = 4b2(1 – e2)
⇒ 1 = 4(1 – e2)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the required value of eccentricity isNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

Q.12. Given the ellipse with equation 9x2 + 25y2 = 225, find the eccentricity and foci.

Ans. Given equation of ellipse is
9x2 + 25y2 = 225
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Here a = 5 and b = 3
b2 = a2(1 – e2)
⇒ 9 = 25(1 – e2)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Now foci = (± ae, 0) =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, eccentricity =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

Q.13. If the eccentricity of an ellipse is 5/8 and the distance between its foci is 10, then find latus rectum of the ellipse.
Ans. Equation of an ellipse isNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Eccentricity,NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Distance between its foci = ae + ae = 2ae
∴ 2ae = 10
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Now b2 = a2(1 – e2)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
So, the length of the latus rectum = NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the length of the latus rectum = 39/4.

Q.14. Find the equation of ellipse whose eccentricity is 2/3, latus rectum is 5 and the centre is (0, 0).
Ans. Equations of ellipse isNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev  ....(i)
Given that,e = 2/3
and latus rectum NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev     ...(ii)
We know that b2 = a2(1 – e2)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
andNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the required equation of ellipse is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

Q.15. Find the distance between the directrices of the ellipseNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Ans. Given equation of ellipse is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Here a2 = 36 ⇒ a = 6
b2 = 20 ⇒ b = 2 √5
We know that b2 = a2(1 – e2)   
⇒ 20 = 36(1 – e2)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Now distance between the directrices is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the required distance = 18.

Q.16. Find the coordinates of a point on the parabola y2 = 8x whose focal distance is 4.
Ans. Given parabola is y2 = 8x     ...(i)
Comparing with the equation of parabola y2 = 4ax
4a = 8 ⇒ a = 2
Now focal distance =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ (x + a) = ± 4
⇒ x + 2 = ± 4
⇒ x = 4 – 2 = 2 and x = – 6
But x ≠ - 6 ∴ x = 2
Put x = 2 in equation (i) we get y2 = 8 x 2 = 16
∴ y = ± 4
So, the coordinates of the point are (2, 4), (2, – 4).
Hence, the required coordinates are (2, 4) and (2, – 4).

Q.17. Find the length of the line-segment joining the vertex of the parabola y2 = 4ax and a point on the parabola where the line-segment makes an angle θ to the xaxis.

Ans. Equation of parabola is y2 = 4ax
Let P(at2, 2at) be any point on the parabola.
In ΔPOA, we have
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ t = 2 cot θ     ...(i)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
[∴ t = 2 cot θ]
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the required length =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

Q.18. If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.

Ans. Given that: Vertex = (0, 4) and Focus = (0, 2)
Let P(x, y) be any point on the parabola. PB is perpendicular to the directrix.
According to the definition of parabola, we have
PF = PB
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
[Equation of directrix is y = 6]
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Squaring both sides, we have
x2 + (y – 2)2 = (y – 6)2
⇒ x2 + y2 + 4 – 4y = y2 + 36 – 12y
⇒ x2 – 4y + 12y – 32 = 0
⇒ x2 + 8y – 32 = 0
Hence, the required equation is x2 + 8y = 32.

Q.19. If the line y = mx + 1 is tangent to the parabola y2 = 4x then find the value of m. 

[Hint: Solving the equation of line and parabola, we obtain a quadratic equation and then apply the tangency condition giving the value of m]
Ans. Given that y2 = 4x     ...(i)
and y = mx + 1    ...(ii)
From eq. (i) and (ii) we get
(mx + 1)2 = 4x
⇒ m2x2 + 1 + 2mx – 4x = 0
 m2x2 + (2m – 4)x + 1 = 0
Applying condition of tangency, we have
(2m – 4)2 – 4m2 x 1 = 0
⇒ 4m2 + 16 - 16m - 4m2 = 0
⇒ – 16m = – 16
⇒ m = 1
Hence, the required value of m is 1.

Q.20. If the distance between the foci of a hyperbola is 16 and its eccentricity is √2 , then obtain the equation of the hyperbola.
Ans. Equation of hyperbola isNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Distance between the foci = 2ae
2ae = 16 ⇒ ae = 8
⇒ a x √2 = 8
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Now, b2 = a2(e2 – 1) [for hyperbola]
⇒ b2 = (4 √2)2 (2 - 1)
⇒ b2 = 32
a = 4√2 ⇒ a= 32
Hence, the required equation isNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ x2 – y2 = 32.

Q.21. Find the eccentricity of the hyperbola 9y2 – 4x2 = 36.
Ans. Given equation is 9y2 – 4x2 = 36
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Clearly it is a vertical hyperbola .
Where a = 3 and b = 2
We know that b2 = a2(e2 – 1)    
⇒ 4 = 9(e2 – 1)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the required value of e isNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

Q.22. Find the equation of the hyperbola with eccentricity 3/2 and foci at (± 2, 0).
Ans. Given that e = 3/2 and foci = (± 2, 0)
We know that foci = (± ae, 0)
∴ ae = 2
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
We know that b2 = a2(e2 – 1)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
So, the equation of the hyperbola is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the required equation isNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

LONG ANSWER TYPE QUESTIONS

Q.23. If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.
Ans. We know that the intersection point of the diameter gives the centre of the circle.
Given equations of diameters are
2x – 3y = 5     ...(i)
3x – 4y = 7     ...(ii)
From eq. (i) we haveNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev   ...(iii)
Putting the value of x in eq. (ii) we have
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ 15 + 9y – 8y = 14
⇒ y = 14 – 15 ⇒ y = – 1
Now from eq. (iii) we have
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
So, the centre of the circle = (1, – 1)
Given that area of the circle = 154    
⇒ πr2 = 154
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ r2 = 7 x 7
⇒ r = 7
So, the equation of the circle is
(x – 1)2 + (y + 1)2 = (7)2
⇒ x2 + 1 – 2x + y2 + 1 + 2y = 49
⇒ x2 + y2 – 2x + 2y = 47
Hence, the required equation of the circle is
x2 + y2 – 2x + 2y = 47

Q.24. Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0.
Ans. Let the equation of the circle be
(x – h)2 + (y – k)2 = r2     ...(i)
If the circle passes through (2, 3) and (4, 5) then
(2 – h)2 + (3 – k)2 = r2      ...(ii)
and (4 – h)2 + (5 – k)2 = r2       ...(iii)
Subtracting eq. (iii) from eq. (ii) we have
(2 – h)2 – (4 – h)2 + (3 – k)2 – (5 – k)2 = 0
⇒ 4 + h2 - 4h - 16 - h2 + 8h + 9 + k2 - 6k - 25 - k2 + 10k = 0
⇒ 4h + 4k – 28 = 0
⇒ h + k = 7      ....(iv)
Since, the centre (h, k) lies on the line y – 4x + 3 = 0
then k – 4h + 3 = 0
⇒ k = 4h – 3
Putting the value of k in eq. (iv) we get
h + 4h – 3 = 7
⇒ 5h = 10 ⇒ h = 2
From (iv) we get k = 5
Putting the value of h and k in eq. (ii) we have
(2 – 2)2 + (3 – 5)2 = r2
⇒ r2 = 4
So, the equation of the circle is
(x – 2)2 + (y – 5)2 = 4
 x2 + 4 – 4x + y2 + 25 – 10y = 4
⇒ x2 + y2 – 4x – 10y + 25 = 0
Hence, the required equation is
x2 + y2 – 4x – 10y + 25 = 0.

Q.25. Find the equation of a circle whose centre is (3, –1) and which cuts off a chord of length 6 units on the line 2x – 5y + 18 = 0. 

[Hint: To determine the radius of the circle, find the perpendicular distance from the centre to the given line.]
Ans. Given that:
Centre of the circle = (3, – 1)
Length of chord AB = 6 units
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Now AB = 6 units.
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
In ΔCPA, AC2 = CP2 + AP2 
= (√29)2 + (3)2 = 29 + 9 = 38
∴ AC = √38
So, the radius of the circle, r = √38
∴ Equation of the circle is
(x – 3)2 + (y + 1)2 = (√38)2  
⇒ (x – 3)2 + (y + 1)2 = 38
⇒ x2 + 9  – 6x + y2 + 1 + 2y = 38
⇒ x2 + y2 – 6x + 2y = 28
Hence, the required equation is x2 + y2 – 6x + 2y = 28.

Q.26. Find the equation of a circle of radius 5 which is touching another circle x2 + y2 – 2x – 4y – 20 = 0 at (5, 5).
Ans. Given circle is
x2 + y2 – 2x – 4y – 20 = 0
2g = – 2 ⇒ g = – 1
2f = – 4 ⇒ f = – 2
∴ Centre C1 = (1, 2)
andNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Let the centre of the required circle be (h, k).
Clearly, P is the mid-point of C1C2
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Radius of the required circle = 5
∴ Eq. of the circle is (x – 9)2 + (y – 8)2 = (5)2 
⇒ x2 + 81 – 18x + y2 + 64 – 16y = 25
⇒ x2 + y2 – 18x – 16y + 145 – 25 = 0
⇒ x2 + y2 – 18x – 16y + 120 = 0
Hence, the required equation is x2 + y2 – 18x – 16y + 120 = 0.

Q.27. Find the equation of a circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x – 1.

Ans. Let the equation of the circle be
(x – h)2 + (y – k)2 = r2
If it passes through (7, 3) then
(7 – h)2 + (3 – k)2 = (3)2     [∵ r = 3]
⇒ 49 + h2 – 14h + 9 + k2 – 6k = 9
⇒ h2 + k2 – 14h – 6k + 49 = 0    ...(i)
If centre (h, k) lies on the line y = x – 1 then
k = h – 1     ...(ii)
Putting the value of k in eq. (i) we get
h2 + (h – 1)2 – 14h – 6(h – 1) + 49 = 0
⇒ h2 + h2 + 1 – 2h – 14h – 6h + 6 + 49 = 0
⇒ 2h2 – 22h + 56 = 0
⇒ h2 – 11h + 28 = 0
⇒ h2 – 7h – 4h + 28 = 0
⇒ h(h – 7) – 4(h – 7) = 0
⇒ (h – 4)(h – 7) = 0
∴ h = 4, h = 7
From eq. (ii) we get k = 4 – 1 = 3 and k = 7 – 1 = 6.
So, the centres are (4, 3) and (7, 6).
∴ Equation of the circle is
Taking centre (4, 3),
(x – 4)2 + (y – 3)2 = 9
x2 + 16 – 8x + y2 + 9 – 6y = 9
⇒ x2 + y2 – 8x – 6y + 16 = 0
Taking centre (7, 6)
(x – 7)2 + (y – 6)2 = 9
⇒ x2 + 49 – 14x + y2 + 36 – 12y = 9
⇒ x2 + y2 – 14x – 12y + 76 = 0
Hence, the required equations are
x2 + y2 – 8x – 6y + 16 = 0
and x2 + y2 – 14x – 12y + 76 = 0.

Q.28. Find the equation of each of the following parabolas 

(a) Directrix x = 0, focus at (6, 0) 
(b) Vertex at (0, 4), focus at (0, 2) 
(c) Focus at (–1, –2), directrix x – 2y + 3 = 0
Ans.
(a) Given that directrix = 0 and focus (6, 0)
∴ The equation of the parabola is
(x – 6)2 + y2 = x2
  x2 + 36 – 12x + y2 = x2
⇒ y2 – 12x + 36 = 0
Hence, the required equations is y2 – 12x + 36 = 0
(b) Given that vertex at (0, 4) and focus at (0, 2).
So, the equation of directrix is y – 6 = 0
According to the definition of the parabola
PF = PM.
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Squaring both the sides, we get
x2 + y2 + 4 – 4y = y2 + 36 – 12y
 x2 + 4 – 4y = 36 – 12y
⇒ x2 + 8y – 32 = 0
⇒ x2 = 32 – 8y
Hence, the required equation is x2 = 32 – 8y.
(c) Given that focus at (– 1, – 2) and directrix x – 2y + 3 = 0
Let (x, y) be any point on the parabola.
According to the definition of the parabola, we have
PF = PM
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Squaring both sides, we get
x2 + 1 + 2x + y2 + 4 + 4y =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ 5x2 + 5 + 10x + 5y2 + 20 + 20y
= x2 + 4y2 + 9 – 4xy – 12y + 6x
 4x2 + y2 + 4xy + 4x + 32y + 16 = 0
Hence, the required equation is
4x2 + 4xy + y2 + 4x + 32y + 16 = 0

Q.29. Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12.

Ans. Let (x, y) be any point.
Given points are (3, 0) and (9, 0)
According to the question, we have
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Putting x2 + 9 – 6x + y2 = k
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Squaring both sides, we have
⇒ 72 – 12x + k = 144 + k - 24√k    
⇒ 24 √k = 144 – 72 + 12x  
⇒ 24 √k = 72 + 12x    
⇒ 2 √k = 6 + x
Again squaring both sides, we get
4k = 36 + x2 + 12x
Putting the value of k, we have
4(x2 + 9 – 6x + y2) = 36 + x2 + 12x
⇒ 4x2 + 36 – 24x + 4y2 = 36 + x2 + 12x
⇒ 3x2 + 4y2 – 36x = 0
Hence, the required equation is 3x2 + 4y2 – 36x = 0

Q.30. Find the equation of the set of all points whose distance from (0, 4) are 2/3 of their distance from the line y = 9.
Ans.
Let P(x, y) be a point.
According to question, we have
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Squaring both sides, we have
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ 9x2 + 9(y – 4)2 = 4y2 + 324 – 72y
⇒ 9x2 + 9y2 + 144 – 72y = 4y2 + 324 – 72y
⇒ 9x2 + 5y2 + 144 – 324 = 0
⇒ 9x2 + 5y2 – 180 = 0
Hence, the required equation is 9x2 + 5y2 – 180 = 0.

Q.31. Show that the set of all points such that the difference of their distances from (4, 0) and (– 4, 0) is always equal to 2 represent a hyperbola.

Ans. Let P(x, y) be any point.
According to the question, we have
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Putting the x2 + y+ 16 = z ...(i)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Squaring both sides, we get
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Again squaring both sides, we have
z2 + 4 – 4z = z2 – 64x2
⇒ 4 – 4z + 64x2 = 0
Putting the value of z, we have
⇒ 4 – 4(x2 + y2 + 16) + 64x2 = 0
⇒ 4 – 4x2 – 4y2 – 64 + 64x2 = 0
⇒ 60x2 – 4y2 – 60 = 0
⇒ 60x2 – 4y2 = 60
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Which represent a hyperbola. Hence proved.

Q.32. Find the equation of the hyperbola with
(a) Vertices (± 5, 0), foci (± 7, 0)
(b) Vertices (0, ± 7), e = 4/3
(c) Foci (0, ± √10), passing through (2, 3)
Ans. (a) Given that vertices (± 5, 0), foci (± 7, 0)
Vertex of hyperbola = (± a, 0) and foci (± ae, 0)
∴ a = 5 and ae = 7 ⇒ 5 x e = 7 ⇒ e = 7/5
Now b2 = a2 (e– 1)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
The equation of the hyperbola is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
(b) Given that vertices (0, ± 7), e = 4/3
Clearly, the hyperbola is vertical.
Vertices = (± 0, a)
∴ a = 7 and e = 4/3
We know that b2 = a2(e2 – 1)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the equation of the hyperbola is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ 9x2 – 7y2 + 343 = 0
(c) Given that: foci = (0 , ± √10)
∴ ae = √10 ⇒ a2e2 = 10
We know that b2 = a2(e2 – 1)
⇒ b2 = a2e2 – a2
⇒ b2 = 10 – a2
Equation of hyperbola is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
If it passes through the point (2, 3) then
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ 90 – 13a2 = a2(10 – a2)
⇒ 90 – 13a2 = 10a2 – a4
  a4 – 23a2 + 90 = 0
⇒ a4 – 18a2 – 5a2 + 90 = 0
⇒ a2(a2 – 18) – 5(a2 – 18) = 0
⇒ (a2 – 18)(a2 – 5) = 0
⇒ a2 = 18, a2 = 5
∴ b2 = 10 –18 = – 8 and b2 = 10 – 5 = 5
b ≠ – 8   ∴ b2 = 5
Here, the required equation isNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

OBJECTIVE ANSWER TYPE QUESTIONS

Q.33. The line x + 3y = 0 is a diameter of the circle x2 + y2 + 6x + 2y = 0.
Ans. Given equation of the circle is
x2 + y2 + 6x + 2y = 0
Centre is (– 3, – 1)
If x + 3y = 0 is the equation of diameter, then the centre (– 3, – 1) will lie on x + 3y = 0
– 3 + 3(– 1) = 0   
⇒ – 6 ≠ 0
So, x + 3y = 0 is not the diameter of the circle.
Hence, the given statement is False.

Q.34. The shortest distance from the point (2, –7) to the circle x+ y2 – 14x – 10y – 151 = 0 is equal to 5. 

[Hint: The shortest distance is equal to the difference of the radius and the distance between the centre and the given point.]
Ans. Given equation of circle is x2 + y2 – 14x – 10y – 151 = 0
Shortest distance = distance between the point (2, – 7)
and the centre – radius of the circle
Centre of the given circle is
2g = – 14 ⇒ g = – 7
2f = – 10 ⇒ f = – 5
∴ Centre = (– g, – f) = (7, 5)
andNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
∴ Shortest distanceNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the given statement is False.

Q.35. If the line lx + my = 1 is a tangent to the circle x2 + y= a2, then the point (l, m) lies on a circle. 

[Hint: Use that distance from the centre of the circle to the given line is equal to radius of the circle.]
Ans. Given equation of circle is x2 + y= a2
and the tangent is lx + my = 1
Here centre is (0, 0) and radius  = a
If (l, m) lies on the circle
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ l2 + m2 = a2 (which is a circle)
So, the point (l, m) lies on the circle.
Hence, the given statement is True.

Q.36. The point (1, 2) lies inside the circle x2 + y2 – 2x + 6y + 1 = 0.

Ans. Given equation of circle is x2 + y2 – 2x + 6y + 1 = 0
Here 2g = – 2 ⇒ g = – 1
2f = 6 ⇒ f = 3
∴ Centre = (– g, – f) = (1, – 3)
and NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
∴ Distance between the point (1, 2) and the centre (1, – 3)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Here 5 > 3, so the point lies out side the circle.
Hence, the given statement is False.

Q.37. The line lx + my + n = 0 will touch the parabola y2 = 4ax if ln = am2.

Ans. Given equation of parabola is y2 = 4ax ...(i)
and the equation of line is lx + my + n = 0 ...(ii)
From eq. (ii), we have
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Putting the value of y in eq. (i) we get
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ l2x2 + n2 + 2lnx – 4am2x = 0
⇒ l2x2 + (2ln – 4am2)x + n2 = 0
If the line is the tangent to the circle, then
b2 – 4ac = 0
(2ln – 4am2)2 – 4l2n2 = 0
⇒ 4l2n2 + 16a2m4 – 16lnm2a – 4l2n2 = 0
⇒ 16a2m4 – 16lnm2a = 0
⇒ 16am2(am2 – ln) = 0
⇒ am2(am2 – ln) = 0
⇒ am2 ≠ 0 ∴ am2 - ln = 0
∴ ln = am2 
Hence, the given statement is True.

Q.38. If P is a point on the ellipseNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRevwhose foci are S and S′, then PS + PS′ = 8.
Ans. Let P(x1, y1) be a point on the ellipse. foci = (± ae, 0)
Here a2 = 25 ⇒ a = 5
b2 = 16 ⇒ b = 4
b2 = a2(1 – e2)
16 = 25(1 – e2)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
So, the foci are S(3, 0) and S’(- 3, 0).
Since PS + PS’ = 2a = 2 x 5 = 10.
Hence, the given statement is False.


Q.39. The line 2x + 3y = 12 touches the ellipseNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRevat the point (3, 2).
Ans. If line 2x + 3y = 12 touches the ellipseNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRevthen the point (3, 2) satisfies both line and ellipse.
∴ For line 2x + 3y = 12
2(3) + 3(2) = 12
6 + 6 = 12
12 = 12 True
For ellipseNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
1 + 1 = 2
2 = 2 True
Hence, the given statement is True.

Q.40. The locus of the point of intersection of lines √3 x − y − 4 √3k = 0 and

√3kx + ky – 4√3 = 0 for different value of k is a hyperbola whose eccentricity is 2. 
[Hint:Eliminate k between the given equations]
Ans. The given equations are
√3x - y - 4 √3k = 0    ...(i)
and √3kx + ky - 4 √3 = 0    ...(ii)
From eq. (i) we get
4√3k = √3x - y
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Putting the value of k in eq. (ii), we get
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ 3x2 - √3xy + √3xy - y2 - 48 = 0     
⇒ 3x2 – y2 = 48
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRevwhich is a hyperbola.
Here a2 = 16, b2 = 48
We know that b2 = a2(e2 – 1)
⇒ 48 = 16(e2 – 1)
⇒ 3 = e2 – 1
⇒ e2 = 4 ⇒ e = 2
Hence, the given statement is True.

FILL IN THE BLANK 

Q.41. The equation of the circle having centre at (3, – 4) and touching the line 5x + 12y – 12 = 0 is  _______. 
[Hint: To determine radius find the perpendicular distance from the centre of the circle to the line.]
Ans. Given equation of the line is 5x + 12y – 12 = 0 and the centre is (3, – 4)
CP = radius of the circle
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
So, the equation of the circle is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the value of the filler is (x – 3)2 + (y + 4)2 = NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

Q.42. The equation of the circle circumscribing the triangle whose sides are the lines y = x + 2, 3y = 4x, 2y = 3x is _______ .

Ans. Let AB represents 2y = 3x     ...(i)
BC represents 3y = 4x    ...(ii)
and AC represents y = x + 2    ...(iii)
From eq. (i) and (ii)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Putting the value of y in eq. (ii) we get
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ 9x = 8x
⇒ x = 0 and y = 0
∴ Coordinates of B = (0, 0)
From eq. (i) and (iii) we get
y = x + 2
Putting y = x + 2 in eq. (i) we get
2(x + 2) = 3x
⇒ 2x + 4 = 3x
⇒ x = 4 and y = 6
∴ Coordinates of A = (4, 6)
Solving eq. (ii) and (iii) we get
y = x + 2
Putting the value of y in eq. (ii) we get
3(x + 2) = 4x ⇒ 3x + 6 = 4x ⇒ x = 6 and y = 8
∴ Coordinates of C = (6, 8)
It implies that the circle is passing through (0, 0), (4, 6) and (6, 8).
We know that the general equation of the circle is
x2 + y2 + 2gx + 2fy + c = 0     ...(i)
Since the points (0, 0), (4, 6) and (6, 8) lie on the circle then
0 + 0 + 0 + 0 + c = 0 ⇒ c = 0
⇒ 16 + 36 + 8g + 12f + c = 0
⇒ 8g + 12f + 0 = – 52
⇒ 2g + 3f = – 13    ...(ii)
and 36 + 64 + 12g + 16f + c = 0
⇒ 12g + 16f + 0 = – 100
⇒ 3g + 4f = – 25 ...(iii)
Solving eq. (ii) and (iii) we get
2g + 3f = – 13
3g + 4f = – 25
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Putting the value of f in eq. (ii) we get
2g + 3 x 11 = - 13
⇒ 2g + 33 = – 13
⇒ 2g = – 46 ⇒ g = – 23
Putting the values of g, f and c in eq. (i) we get
x2 + y2 + 2(– 23)x + 2(11)y + 0 = 0
⇒ x2 + y– 46x + 22y = 0
Hence, the value of the filler is x2 + y2 – 46x + 22y = 0.

Q.43. An ellipse is described by using an endless string which is passed over two pins. If the axes are 6 cm and 4 cm, the length of the string and distance between the pins are _______.
Ans. Let equation of ellipse isNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Here 2a = 6 ⇒ a = 3
and 2b = 4 ⇒ b = 2
We know that c2 = a2 – b2
= (3)2 – (2)2 = 9 – 4 = 5
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Length of string = 2a + 2ae = 2a(1 + e)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Distance between the pins = CC’ = 2ae =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence the value of the filler are 6 + 2√5 cm and 2√5 cm.

Q.44. The equation of the ellipse having foci (0, 1), (0, –1) and minor axis of length 1 is_______.

Ans. We know that the foci of the ellipse are (0, ± ae) and given foci are (0, ± 1), so ae = 1
Length of minor axis = 2b = 1 ⇒NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
We know that b2 = a2(1 – e2)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
∴ Equation of ellipse is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the value of the filler isNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

Q.45. The equation of the parabola having focus at (–1, –2) and the directrix x – 2y + 3 = 0 is ________.
Ans. Let (x1, y1) be any point on the parabola.
According to the definition of the parabola
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Squaring both sides, we get
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ 5x12 + 5y12 + 10x1 + 20 y1 + 25
= x21 + 4 y12 - 4x1 y1 - 12 y1 + 6x1 + 9
 4 x12 + y12 + 4 x1 + 32 y1 + 4 x1 y1 + 16 = 0
Hence, the value of the filler is 4x2 + 4xy + y2 + 4x + 32y + 16 = 0.

Q.46. The equation of the hyperbola with vertices at (0, ± 6) and eccentricity 5/3 is________and its foci are _______ .
Ans. Let equation of the hyperbola isNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Vertices are (0, ± b) ∴ b = 6 and e = 5/3
We know thatNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ a2 = 64
So the equation of the hyperbola is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
and foci = (0, ± be) =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the value of the filler isNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev

Choose the correct answer out of the given four options (M.C.Q.)
Q.47. The area of the circle centred at (1, 2) and passing through (4, 6) is 
(a) 5 π 
(b) 10π 
(c) 25π 
(d) None of these
Ans. (c)
Solution.

Given that the centre of the circle is (1, 2)
Radius of the circle =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
So, the area of the circle = πr2
= π x (5)2 = 25π
Hence, the correct option is (c).

Q.48. Equation of a circle which passes through (3, 6) and touches the axes is 
(a) x2 + y2 + 6x + 6y + 3 = 0 
(b) x2 + y2 – 6x – 6y – 9 = 0 
(c) x2 + y2 – 6x – 6y + 9 = 0 
(d) None of these
Ans. (c)
Solution.

Let the required circle touch the axes at (a, 0) and (0, a)
∴ Centre is (a, a) and r = a
So the equation of the circle is (x – a)2 + (y – a)2 = a2
If it passes through a point P(3, 6) then
(3 – a)2 + (6 – a)2 = a2
 9 + a2 – 6a + 36 + a2 – 12a = a2     
⇒ a2 – 18a + 45 = 0
⇒ a2 – 15a – 3a + 45 = 0
⇒ a(a – 15) – 3(a – 15) = 0
⇒ (a – 3) (a – 15) = 0
   a = 3 and a = 15 which is not possible
∴ a = 3
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
So, the required equation of the circle is
(x – 3)2 + (y – 3)2 = 9
⇒ x2 + 9 – 6x + y2 + 9 – 6y = 9
⇒ x2 + y2 – 6x – 6y + 9 = 0
Hence, the correct option is (c).

Q.49. Equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is 
(a) x2 + y2 + 13y = 0 
(b) 3x2 + 3y2 + 13x + 3 = 0 
(c) 6x2 + 6y2 – 13x = 0 
(d) x2 + y2 + 13x + 3 = 0
Ans. (a)
Solution.

Let the equation of the circle be
(x – h)2 + (y – k)2 = r2
Let the centre be (0, a)
∴ Radius r = a
So, the equation of the circle is
(x – 0)2 + (y – a)2 = a2
⇒ x2 + (y – a)2 = a2    
⇒ x2 + y2 + a2 – 2ay = a2
⇒ x2 + y2 – 2ay = 0     ...(i)
Now CP = r
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ 13 + a2 – 6a = a2
⇒ 13 – 6a = 0
∴ a = 13/6
Putting the value of a in eq. (i) we get
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev   
⇒ 3x2 + 3y2 – 13y = 0
(Note: (a) option is correct and it should be 3x2 + 3y2 – 13y = 0)
Hence, the correct option is (a).

Q.50.
 The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
(a) x2 + y2 = 9a2 
(b) x2 + y2 = 16a2 
(c) x2 + y2 = 4a2 
(d) x2 + y2 = a2 
[Hint: Centroid of the triangle coincides with the centre of the circle and the 
radius of the circle is 2/3 of the length of the median]
Ans. (c)
Solution.

Let ABC be an equilateral triangle in which median AD = 3a.
Centre of the circle is same as the centroid of the triangle i.e., (0, 0)
AG : GD = 2 : 1
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
So, NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
∴ The equation of the circle is
(x – 0)2 + (y – 0)2 = (2a)    
⇒ x2 + y= 4a2 
Hence, the correct option is (c).

Q.51. If the focus of a parabola is (0, –3) and its directrix is y = 3, then its equation is 

(a) x2 = –12y 
(b) x2 = 12y 
(c) y2 = –12x 
(d) y2 = 12x
Ans. (a)
Solution.

According to the definition of parabola
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Squaring both sides, we have
x2 + y2 + 9 + 6y = y2 + 9 – 6y
 x2 + 9 + 6y = 9 – 6y     
⇒ x2 = – 12y
Hence, the correct option is (a).

Q.52. If the parabola y2 = 4ax passes through the point (3, 2), then the length of its latus rectum is

(a) 2/3
(b) 4/3
(c) 1/3
(d) 4
Ans. (b)
Solution.

Given parabola is y2 = 4ax
If the parabola is passing through (3, 2) then
(2)2 = 4a x 3
⇒ 4 = 12a ⇒ a = 1/3
Now length of the latus rectum = 4a =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the correct option is (b).

Q.53. If the vertex of the parabola is the point (– 3, 0) and the directrix is the line x + 5 = 0, then the equation is 
(a) y2 = 8(x + 3) 
(b) x2 = 8(y + 3) 
(c) y2 = – 8(x + 3) 
(d) y2 = 8(x + 5)
Ans. (a)
Solution.

Given that vertex = (– 3, 0)
∴ a = – 3
and directrix is x + 5 = 0
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
According to the definition of the parabola, we get
AF = AD i.e., A is the mid-point of DF
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
and NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
∴ Focus F = (– 1, 0)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Squaring both sides, we get
(x + 1)2 + y2 = (x + 5)2
 x2 + 1 + 2x + y2 = x2 + 25 + 10x
⇒ y2 = 10x – 2x + 24 ⇒  y2 = 8x + 24
⇒ y2 = 8(x + 3)
Hence, the correct option is (a).

Q.54. The equation of the ellipse, whose focus is (1, – 1), the directrix the line x – y – 3 = 0 and eccentricity 1/2, is

(a) 7x+ 2xy + 7y2 – 10x + 10y + 7 = 0
(b) 7x2 + 2xy + 7y2 + 7 = 0
(c) 7x2 + 2xy + 7y2 + 10x – 10y – 7 = 0
(d) None of the above

Ans. (a)
Solution.

Given that focus of the ellipse is (1, – 1) and the equation of the directrix is x – y – 3 = 0 and e = 1/2.
Let P(x, y) by any point on the parabola
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Squaring both sides, we have
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
⇒ 8x2 + 8y2 – 16x + 16y + 16 = x+ y2 – 2xy + 6y – 6x + 9
⇒ 7x2 + 7y2 + 2xy – 10x + 10y + 7 = 0
Hence, the correct option is (a).

Q.55. The length of the latus rectum of the ellipse 3x2 + y2 = 12 is
(a) 4 
(b) 3 
(c) 8 
(d) 4√3
Ans. (d)
Solution.

Equation of the ellipse is
3x2 + y= 12
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Here a2 = 4 ⇒ a = 2
b2 = 12 ⇒ b = 2√ 3
Length of the latus rectum =NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the correct option is (d).

Q.56. If e is the eccentricity of the ellipseNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev then

(a) b2 = a2(1 – e2
(b) a2 = b2(1 – e2
(c) a2 = b2(e2 – 1) 
(d) b2 = a2(e2 – 1)
Ans. (b)
Solution.

Given equation isNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
∴Eccentricity eNCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the correct option is (b).

Q.57. The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half of the distance between the foci is
(a) 4/3
(b) 4/√3
(c) 2/√3
(d) None of these
Ans. (c)
Solution.

Length of the latus rectum of the hyperbola
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Distance between the foci = 2ae
Transverse axis = 2a
and Conjugate axis = 2b
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev[from eq. (i)]
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Now b= a2(e2  – 1)
⇒ 4a = a2(e2  – 1)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the correct option is (c).

Q.58. The distance between the foci of a hyperbola is 16 and its eccentricity is 2 . Its equation is
(a) x2 – y2 = 32
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
(c) 2x2 – 3y2 = 7 
(d) None of these
Ans. (a)
Solution.

We know that the distance between the foci = 2ae
∴ 2ae = 16 ⇒ ae = 8
Given that e = √2
∴ 2a = 8 ⇒ a = 4 √2
Now b2 = a2(e2 – 1)
⇒ b2 = 32(2 – 1)
⇒ b2 = 32
So, the equation of the hyperbola is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the correct option is (a).

Q.59. Equation of the hyperbola with eccentricty 3/2 and foci at (± 2, 0) is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
(d) None of these
Ans. (a)
Solution.

Given that e = 3/2
and foci = (± ae, 0) = (± 2, 0)
∴ ae = 2
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Now we know that b2 = a2(e2 – 1)
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
So, the equation of the hyperbola is
NCERT Exemplar - Circle, Parabola, Ellipse & Hyperbola Notes | EduRev
Hence, the correct option is (a).

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