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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 â€“ h)^{2} + (y â€“ k)^{2} = r^{2}

So, the equation of the required circle

â‡’ (x â€“ a)^{2} + (y â€“ a)^{2} = a^{2}

â‡’ x^{2} â€“ 2ax + a^{2} + y^{2} â€“ 2ay + a^{2} = a^{2}

â‡’ x^{2} + y^{2} â€“ 2ax â€“ 2ay + a^{2} = 0

Hence, the required equation is

x^{2} + y^{2} â€“ 2ax â€“ 2ay + a^{2} = 0**Q.2. Show that the point (x, y) given by x = ****lies on a circle for all real values of t such that â€“1 â‰¤ t â‰¤ 1 where a is any given real numbers.****Ans.**

Given

= a^{2}

âˆ´ x^{2} + y^{2} = a^{2} 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 x^{2} + y^{2} + 2gx + 2fy + c = 0

where the centre is (â€“ g, â€“ f) and radius =

If it passes through (0, 0)

âˆ´ c = 0

If it passes through (a, 0) and (0, b) then

a^{2} + 2ga + c = 0 â‡’ a^{2} + 2ga = 0 [âˆµ c = 0]

âˆ´

and 0 + b^{2} + 0 + 2fb + c = 0 â‡’ b^{2} + 2fb = 0 [âˆµ c = 0]

â‡’

Hence, the coordinates of centre of the circle are (â€“ g, â€“ f)**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

âˆ´ r = 2

So, the equation of the circle is

(x â€“ h)^{2} + (y â€“ k)^{2} = r^{2}

â‡’ (x â€“ 1)^{2} + (y â€“ 2)^{2} = (2)^{2}

â‡’ x^{2} â€“ 2x + 1 + y^{2} â€“ 4y + 4 = 4

â‡’ x^{2} + y^{2} â€“ 2x â€“ 4y + 1 = 0

Hence, the required equation is

x^{2} + y^{2} â€“ 2x â€“ 4y + 1 = 0.

Q.5. If the lines 3x â€“ 4y + 4 = 0 and 6x â€“ 8y â€“ 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 â€“ 4y + 4 = 0

and 6x â€“ 8y â€“ 7 = 0 â‡’

Sincethen the lines are parallel.

So, the distance between the parallel lines

Diameter = 3/2

âˆ´ 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 â€“ 4y + 8 = 0 and lies in the third quadrant. ****[Hint: Let a be the radius of the circle, then (â€“ a, â€“ 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 = (â€“ a, â€“ a)

Distance of the line 3x â€“ 4y + 8 = 0

From the centre = Radius of the circle

â‡’

â‡’ a = 5a â€“ 8

â‡’ 5a â€“ a = 8

â‡’ 4a = 8 â‡’ a = 2

and

âˆ´ The equation of the circle is

(x + 2)^{2} + (y + 2)^{2} = (2)^{2}

â‡’ x^{2} + 4x + 4 + y^{2} + 4y + 4 = 4

â‡’ x^{2} + y^{2} + 4x + 4y + 4 = 0

Hence, the required equation of the circle

x^{2} + y^{2} + 4x + 4y + 4 = 0.**Q.7. If one end of a diameter of the circle x ^{2} + y^{2} â€“ 4x â€“ 6y + 11 = 0 is (3, 4), then find the coordinate of the other end of the diameter.**

Equation of given circle is

x

Centre = (â€“ g, â€“ f) = (2, 3)

Hence, the required coordinates are (1, 2).

3x + y = 14 ...(i)

and 2x + 5y = 18 ...(ii)

From eq. (i) we get y = 14 â€“ 3x ...(iii)

Putting the value of y in eq. (ii) we get

â‡’ 2x + 5(14 â€“ 3x) = 18

â‡’ 2x + 70 â€“ 15x = 18

â‡’ â€“ 13x = â€“ 70 + 18

â‡’ â€“ 13x = â€“ 52

âˆ´ x = 4

From eq. (iii) we get,

y = 14 â€“ 3 x 4 = 2

âˆ´ Point of intersection is (4, 2)

Now, radius r =

So, the equation of circle is (x â€“ h)

â‡’ (x â€“ 1)

â‡’ x

â‡’ x

Hence, the required equation is x

Centre = (0, 0)

radius r = 4

Perpendicular from the origin to the given line y = âˆš3x + k is equal to the radius.

Hence, the required values of k are Â± 8.

x

Centre = (â€“ g, â€“ f) = (3, â€“ 6)

Since the circle is concentric with the given circle

âˆ´ Centre = (3, â€“ 6)

Now let the radius of the circle is r

âˆ´

Area of the given circle (i) = Ï€r

Area of the required circle = 2 x 30Ï€ = 60Ï€ sq. unit

If r

Ï€r

So, the required equations of the circle is

(x - 3)

â‡’ x

â‡’ x

Hence, the required equation is x

Length of major axis = 2a

Length of minor axis = 2b.

and the length of latus rectum =

According to the question, we have

Now b

where e is the eccentricity

â‡’ b

â‡’ 1 = 4(1 â€“ e

Hence, the required value of eccentricity is

Q.12. Given the ellipse with equation 9x

9x

â‡’

â‡’

Here a = 5 and b = 3

b

â‡’ 9 = 25(1 â€“ e

â‡’

Now foci = (Â± ae, 0) =

Hence, eccentricity =

Eccentricity,

Distance between its foci = ae + ae = 2ae

âˆ´ 2ae = 10

â‡’

Now b

â‡’

â‡’

So, the length of the latus rectum =

Hence, the length of the latus rectum = 39/4.

Given that,e = 2/3

and latus rectum

â‡’ ...(ii)

We know that b

â‡’

â‡’

and

Hence, the required equation of ellipse is

Here a

b

We know that b

â‡’ 20 = 36(1 â€“ e

â‡’

â‡’

â‡’

Now distance between the directrices is

Hence, the required distance = 18.

Comparing with the equation of parabola y

4a = 8 â‡’ a = 2

Now focal distance =

â‡’

â‡’ (x + a) = Â± 4

â‡’ x + 2 = Â± 4

â‡’ x = 4 â€“ 2 = 2 and x = â€“ 6

But x â‰ - 6 âˆ´ x = 2

Put x = 2 in equation (i) we get y

âˆ´ y = Â± 4

So, the coordinates of the point are (2, 4), (2, â€“ 4).

Hence, the required coordinates are (2, 4) and (2, â€“ 4).

Q.17. Find the length of the line-segment joining the vertex of the parabola y

Let P(at

In Î”POA, we have

â‡’ t = 2 cot Î¸ ...(i)

[âˆ´ t = 2 cot Î¸]

Hence, the required length =

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.

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

[Equation of directrix is y = 6]

Squaring both sides, we have

x

â‡’ x

â‡’ x

â‡’ x

Hence, the required equation is x

Q.19. If the line y = mx + 1 is tangent to the parabola y

and y = mx + 1 ...(ii)

From eq. (i) and (ii) we get

(mx + 1)

â‡’ m

â‡’ m

Applying condition of tangency, we have

(2m â€“ 4)

â‡’ 4m

â‡’ â€“ 16m = â€“ 16

â‡’ m = 1

Hence, the required value of m is 1.

Distance between the foci = 2ae

2ae = 16 â‡’ ae = 8

â‡’ a x âˆš2 = 8

â‡’

Now, b

â‡’ b

â‡’ b

a = 4âˆš2 â‡’ a

Hence, the required equation is

â‡’ x

â‡’

Clearly it is a vertical hyperbola .

Where a = 3 and b = 2

We know that b

â‡’ 4 = 9(e

â‡’

â‡’

âˆ´

Hence, the required value of e is

We know that foci = (Â± ae, 0)

âˆ´ ae = 2

â‡’

â‡’

We know that b

â‡’

So, the equation of the hyperbola is

â‡’

â‡’

Hence, the required equation is

**LONG ANSWER TYPE QUESTIONS**

**Q.23. If the lines 2x â€“ 3y = 5 and 3x â€“ 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 â€“ 3y = 5 ...(i)

3x â€“ 4y = 7 ...(ii)

From eq. (i) we have ...(iii)

Putting the value of x in eq. (ii) we have

â‡’ 15 + 9y â€“ 8y = 14

â‡’ y = 14 â€“ 15 â‡’ y = â€“ 1

Now from eq. (iii) we have

So, the centre of the circle = (1, â€“ 1)

Given that area of the circle = 154

â‡’ Ï€r^{2} = 154

â‡’

â‡’ r^{2} = 7 x 7

â‡’ r = 7

So, the equation of the circle is

(x â€“ 1)^{2} + (y + 1)^{2} = (7)^{2}

â‡’ x^{2} + 1 â€“ 2x + y^{2} + 1 + 2y = 49

â‡’ x^{2} + y^{2} â€“ 2x + 2y = 47

Hence, the required equation of the circle is

x^{2} + y^{2} â€“ 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 â€“ 4x + 3 = 0.****Ans. **Let the equation of the circle be

(x â€“ h)^{2} + (y â€“ k)^{2} = r^{2} ...(i)

If the circle passes through (2, 3) and (4, 5) then

(2 â€“ h)^{2} + (3 â€“ k)^{2} = r^{2} ...(ii)

and (4 â€“ h)^{2} + (5 â€“ k)^{2} = r^{2} ...(iii)

Subtracting eq. (iii) from eq. (ii) we have

(2 â€“ h)^{2} â€“ (4 â€“ h)^{2} + (3 â€“ k)^{2} â€“ (5 â€“ k)^{2} = 0

â‡’ 4 + h^{2} - 4h - 16 - h^{2} + 8h + 9 + k^{2} - 6k - 25 - k^{2} + 10k = 0

â‡’ 4h + 4k â€“ 28 = 0

â‡’ h + k = 7 ....(iv)

Since, the centre (h, k) lies on the line y â€“ 4x + 3 = 0

then k â€“ 4h + 3 = 0

â‡’ k = 4h â€“ 3

Putting the value of k in eq. (iv) we get

h + 4h â€“ 3 = 7

â‡’ 5h = 10 â‡’ h = 2

From (iv) we get k = 5

Putting the value of h and k in eq. (ii) we have

(2 â€“ 2)^{2} + (3 â€“ 5)^{2} = r^{2}

â‡’ r^{2} = 4

So, the equation of the circle is

(x â€“ 2)^{2} + (y â€“ 5)^{2} = 4

â‡’ x^{2} + 4 â€“ 4x + y^{2} + 25 â€“ 10y = 4

â‡’ x^{2} + y^{2} â€“ 4x â€“ 10y + 25 = 0

Hence, the required equation is

x^{2} + y^{2} â€“ 4x â€“ 10y + 25 = 0.

Q.25. Find the equation of a circle whose centre is (3, â€“1) and which cuts off a chord of length 6 units on the line 2x â€“ 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, â€“ 1)

Length of chord AB = 6 units

Now AB = 6 units.

âˆ´

In Î”CPA, AC^{2} = CP^{2} + AP^{2}

= (âˆš29)^{2} + (3)^{2} = 29 + 9 = 38

âˆ´ AC = âˆš38

So, the radius of the circle, r = âˆš38

âˆ´ Equation of the circle is

(x â€“ 3)^{2} + (y + 1)^{2} = (âˆš38)^{2}

â‡’ (x â€“ 3)^{2} + (y + 1)^{2} = 38

â‡’ x^{2} + 9 â€“ 6x + y^{2} + 1 + 2y = 38

â‡’ x^{2} + y^{2} â€“ 6x + 2y = 28

Hence, the required equation is x^{2} + y^{2} â€“ 6x + 2y = 28.**Q.26. Find the equation of a circle of radius 5 which is touching another circle x ^{2} + y^{2} â€“ 2x â€“ 4y â€“ 20 = 0 at (5, 5).**

x

2g = â€“ 2 â‡’ g = â€“ 1

2f = â€“ 4 â‡’ f = â€“ 2

âˆ´ Centre C

and

Let the centre of the required circle be (h, k).

Clearly, P is the mid-point of C

Radius of the required circle = 5

âˆ´ Eq. of the circle is (x â€“ 9)

â‡’ x

â‡’ x

â‡’ x

Hence, the required equation is x

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 â€“ 1.

(x â€“ h)

If it passes through (7, 3) then

(7 â€“ h)

â‡’ 49 + h

â‡’ h

If centre (h, k) lies on the line y = x â€“ 1 then

k = h â€“ 1 ...(ii)

Putting the value of k in eq. (i) we get

h

â‡’ h

â‡’ 2h

â‡’ h

â‡’ h

â‡’ h(h â€“ 7) â€“ 4(h â€“ 7) = 0

â‡’ (h â€“ 4)(h â€“ 7) = 0

âˆ´ h = 4, h = 7

From eq. (ii) we get k = 4 â€“ 1 = 3 and k = 7 â€“ 1 = 6.

So, the centres are (4, 3) and (7, 6).

âˆ´ Equation of the circle is

Taking centre (4, 3),

(x â€“ 4)

x

â‡’ x

Taking centre (7, 6)

(x â€“ 7)

â‡’ x

â‡’ x

Hence, the required equations are

x

and x

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

âˆ´ The equation of the parabola is

(x â€“ 6)

â‡’ x

â‡’ y

Hence, the required equations is y

So, the equation of directrix is y â€“ 6 = 0

According to the definition of the parabola

PF = PM.

Squaring both the sides, we get

x

â‡’ x

â‡’ x

â‡’ x

Hence, the required equation is x

Let (x, y) be any point on the parabola.

According to the definition of the parabola, we have

PF = PM

Squaring both sides, we get

x

â‡’ 5x

= x

â‡’ 4x

Hence, the required equation is

4x

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.

Given points are (3, 0) and (9, 0)

According to the question, we have

Putting x

Squaring both sides, we have

â‡’ 72 â€“ 12x + k = 144 + k - 24âˆšk

â‡’ 24 âˆšk = 144 â€“ 72 + 12x

â‡’ 24 âˆšk = 72 + 12x

â‡’ 2 âˆšk = 6 + x

Again squaring both sides, we get

4k = 36 + x

Putting the value of k, we have

4(x

â‡’ 4x

â‡’ 3x

Hence, the required equation is 3x

Let P(x, y) be a point.

According to question, we have

Squaring both sides, we have

â‡’ 9x

â‡’ 9x

â‡’ 9x

â‡’ 9x

Hence, the required equation is 9x

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

According to the question, we have

Putting the x

Squaring both sides, we get

â‡’

â‡’

â‡’

Again squaring both sides, we have

z

â‡’ 4 â€“ 4z + 64x

Putting the value of z, we have

â‡’ 4 â€“ 4(x

â‡’ 4 â€“ 4x

â‡’ 60x

â‡’ 60x

â‡’

â‡’

Which represent a hyperbola. Hence proved.

Vertex of hyperbola = (Â± a, 0) and foci (Â± ae, 0)

âˆ´ a = 5 and ae = 7 â‡’ 5 x e = 7 â‡’ e = 7/5

Now b

The equation of the hyperbola is

(b) Given that vertices (0, Â± 7), e = 4/3

Clearly, the hyperbola is vertical.

Vertices = (Â± 0, a)

âˆ´ a = 7 and e = 4/3

We know that b

â‡’

â‡’

â‡’

Hence, the equation of the hyperbola is

â‡’ 9x

(c) Given that: foci = (0 , Â± âˆš10)

âˆ´ ae = âˆš10 â‡’ a

We know that b

â‡’ b

â‡’ b

Equation of hyperbola is

â‡’

If it passes through the point (2, 3) then

â‡’

â‡’ 90 â€“ 13a

â‡’ 90 â€“ 13a

â‡’ a

â‡’ a

â‡’ a

â‡’ (a

â‡’ a

âˆ´ b

b â‰ â€“ 8 âˆ´ b

Here, the required equation is

**OBJECTIVE ANSWER TYPE QUESTIONS**

**Q.33. The line x + 3y = 0 is a diameter of the circle x ^{2} + y^{2} + 6x + 2y = 0.**

x

Centre is (â€“ 3, â€“ 1)

If x + 3y = 0 is the equation of diameter, then the centre (â€“ 3, â€“ 1) will lie on x + 3y = 0

â€“ 3 + 3(â€“ 1) = 0

â‡’ â€“ 6 â‰ 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, â€“7) to the circle x

Shortest distance = distance between the point (2, â€“ 7)

and the centre â€“ radius of the circle

Centre of the given circle is

2g = â€“ 14 â‡’ g = â€“ 7

2f = â€“ 10 â‡’ f = â€“ 5

âˆ´ Centre = (â€“ g, â€“ f) = (7, 5)

and

âˆ´ Shortest distance

Hence, the given statement is False.

Q.35. If the line lx + my = 1 is a tangent to the circle x

and the tangent is lx + my = 1

Here centre is (0, 0) and radius = a

If (l, m) lies on the circle

âˆ´

â‡’

â‡’ l

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 x

Here 2g = â€“ 2 â‡’ g = â€“ 1

2f = 6 â‡’ f = 3

âˆ´ Centre = (â€“ g, â€“ f) = (1, â€“ 3)

and

âˆ´ Distance between the point (1, 2) and the centre (1, â€“ 3)

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 y

and the equation of line is lx + my + n = 0 ...(ii)

From eq. (ii), we have

Putting the value of y in eq. (i) we get

â‡’ l

â‡’ l

If the line is the tangent to the circle, then

b

(2ln â€“ 4am

â‡’ 4l

â‡’ 16a

â‡’ 16am

â‡’ am

â‡’ am

âˆ´ ln = am

Hence, the given statement is True.

Here a

b

b

16 = 25(1 â€“ e

â‡’

â‡’

âˆ´

So, the foci are S(3, 0) and Sâ€™(- 3, 0).

Since PS + PSâ€™ = 2a = 2 x 5 = 10.

Hence, the given statement is False.

**Q.39. The line 2x + 3y = 12 touches the ellipseat the point (3, 2).****Ans. **If line 2x + 3y = 12 touches the ellipsethen the point (3, 2) satisfies both line and ellipse.

âˆ´ For line 2x + 3y = 12

2(3) + 3(2) = 12

6 + 6 = 12

12 = 12 True

For ellipse

1 + 1 = 2

2 = 2 True

Hence, the given statement is True.

Q.40. The locus of the point of intersection of lines âˆš3 x âˆ’ y âˆ’ 4 âˆš3k = 0 and**âˆš3kx + ky â€“ 4âˆš3 = 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

âˆš3x - y - 4 âˆš3k = 0 ...(i)

and âˆš3kx + ky - 4 âˆš3 = 0 ...(ii)

From eq. (i) we get

4âˆš3k = âˆš3x - y

âˆ´

Putting the value of k in eq. (ii), we get

â‡’ 3x^{2} - âˆš3xy + âˆš3xy - y^{2} - 48 = 0

â‡’ 3x^{2} â€“ y^{2} = 48

â‡’which is a hyperbola.

Here a^{2} = 16, b^{2} = 48

We know that b^{2} = a^{2}(e^{2} â€“ 1)

â‡’ 48 = 16(e^{2} â€“ 1)

â‡’ 3 = e^{2} â€“ 1

â‡’ e^{2} = 4 â‡’ e = 2

Hence, the given statement is True.

**FILL IN THE BLANK **

**Q.41. The equation of the circle having centre at (3, â€“ 4) and touching the line 5x + 12y â€“ 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 â€“ 12 = 0 and the centre is (3, â€“ 4)

CP = radius of the circle

â‡’

â‡’

â‡’

So, the equation of the circle is

Hence, the value of the filler is (x â€“ 3)^{2} + (y + 4)^{2} =

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)

Putting the value of y in eq. (ii) we get

â‡’ 9x = 8x

â‡’ x = 0 and y = 0

âˆ´ 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

â‡’ 2x + 4 = 3x

â‡’ x = 4 and y = 6

âˆ´ 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 â‡’ 3x + 6 = 4x â‡’ x = 6 and y = 8

âˆ´ 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

x^{2} + y^{2} + 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 â‡’ c = 0

â‡’ 16 + 36 + 8g + 12f + c = 0

â‡’ 8g + 12f + 0 = â€“ 52

â‡’ 2g + 3f = â€“ 13 ...(ii)

and 36 + 64 + 12g + 16f + c = 0

â‡’ 12g + 16f + 0 = â€“ 100

â‡’ 3g + 4f = â€“ 25 ...(iii)

Solving eq. (ii) and (iii) we get

2g + 3f = â€“ 13

3g + 4f = â€“ 25

â‡’

Putting the value of f in eq. (ii) we get

2g + 3 x 11 = - 13

â‡’ 2g + 33 = â€“ 13

â‡’ 2g = â€“ 46 â‡’ g = â€“ 23

Putting the values of g, f and c in eq. (i) we get

x^{2} + y^{2} + 2(â€“ 23)x + 2(11)y + 0 = 0

â‡’ x^{2} + y^{2 }â€“ 46x + 22y = 0

Hence, the value of the filler is x^{2} + y^{2} â€“ 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 is

Here 2a = 6 â‡’ a = 3

and 2b = 4 â‡’ b = 2

We know that c^{2} = a^{2} â€“ b^{2}

= (3)^{2} â€“ (2)^{2} = 9 â€“ 4 = 5

âˆ´

Length of string = 2a + 2ae = 2a(1 + e)

Distance between the pins = CCâ€™ = 2ae =

Hence the value of the filler are 6 + 2âˆš5 cm and 2âˆš5 cm.

Q.44. The equation of the ellipse having foci (0, 1), (0, â€“1) and minor axis of length 1 is_______.**Ans. **We know that the foci of the ellipse are (0, Â± ae) and given foci are (0, Â± 1), so ae = 1

Length of minor axis = 2b = 1 â‡’

We know that b^{2} = a^{2}(1 â€“ e^{2})

â‡’

âˆ´ Equation of ellipse is

â‡’

â‡’

Hence, the value of the filler is**Q.45. The equation of the parabola having focus at (â€“1, â€“2) and the directrix x â€“ 2y + 3 = 0 is ________.****Ans. **Let (x_{1}, y_{1}) be any point on the parabola.

According to the definition of the parabola

Squaring both sides, we get

â‡’ 5x_{1}^{2} + 5y_{1}^{2} + 10x_{1} + 20 y_{1} + 25

= x^{2}_{1} + 4 y_{1}^{2} - 4x_{1} y_{1} - 12 y_{1} + 6x_{1} + 9

â‡’ 4 x_{1}^{2} + y_{1}^{2} + 4 x_{1} + 32 y_{1} + 4 x_{1} y_{1} + 16 = 0

Hence, the value of the filler is 4x^{2} + 4xy + y^{2} + 4x + 32y + 16 = 0.**Q.46. ****The equation of the hyperbola with vertices at (0, Â± 6) and eccentricity 5/3 ****is________and its foci are _______ .****Ans. **Let equation of the hyperbola is

Vertices are (0, Â± b) âˆ´ b = 6 and e = 5/3

We know that

â‡’

â‡’

â‡’ a^{2} = 64

So the equation of the hyperbola is

and foci = (0, Â± be) =

Hence, the value of the filler is**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 Ï€ ****(b) 10Ï€ ****(c) 25Ï€ ****(d) None of these****Ans.** (c)

Solution.

Given that the centre of the circle is (1, 2)

Radius of the circle =

So, the area of the circle = Ï€r^{2}

= Ï€ x (5)^{2} = 25Ï€

Hence, the correct option is (c).**Q.48. Equation of a circle which passes through (3, 6) and touches the axes is ****(a) x ^{2} + y^{2} + 6x + 6y + 3 = 0 **

Solution.

Let the required circle touch the axes at (a, 0) and (0, a)

âˆ´ Centre is (a, a) and r = a

So the equation of the circle is (x â€“ a)

If it passes through a point P(3, 6) then

(3 â€“ a)

â‡’ 9 + a

â‡’ a

â‡’ a

â‡’ a(a â€“ 15) â€“ 3(a â€“ 15) = 0

â‡’ (a â€“ 3) (a â€“ 15) = 0

â‡’ a = 3 and a = 15 which is not possible

âˆ´ a = 3

So, the required equation of the circle is

(x â€“ 3)

â‡’ x

â‡’ x

Hence, the correct option is (c).

Solution.

Let the equation of the circle be

(x â€“ h)

Let the centre be (0, a)

âˆ´ Radius r = a

So, the equation of the circle is

(x â€“ 0)

â‡’ x

â‡’ x

â‡’ x

Now CP = r

â‡’

â‡’

â‡’

â‡’ 13 + a

â‡’ 13 â€“ 6a = 0

âˆ´ a = 13/6

Putting the value of a in eq. (i) we get

â‡’ 3x

(Note: (a) option is correct and it should be 3x

Hence, the correct option is (a).

Q.50.

(a) x

(b) x

(c) x

(d) x

[Hint: Centroid of the triangle coincides with the centre of the circle and the

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

So,

âˆ´ The equation of the circle is

(x â€“ 0)

â‡’ x

Hence, the correct option is (c).

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

Solution.

According to the definition of parabola

Squaring both sides, we have

x

â‡’ x

â‡’ x

Hence, the correct option is (a).

Q.52. If the parabola y

Solution.

Given parabola is y

If the parabola is passing through (3, 2) then

(2)

â‡’ 4 = 12a â‡’ a = 1/3

Now length of the latus rectum = 4a =

Hence, the correct option is (b).

Solution.

Given that vertex = (â€“ 3, 0)

âˆ´ a = â€“ 3

and directrix is x + 5 = 0

According to the definition of the parabola, we get

AF = AD i.e., A is the mid-point of DF

âˆ´

and

âˆ´ Focus F = (â€“ 1, 0)

Squaring both sides, we get

(x + 1)

â‡’ x

â‡’ y

â‡’ y

Hence, the correct option is (a).

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

(b) 7x

(c) 7x

(d) None of the above

Solution.

Given that focus of the ellipse is (1, â€“ 1) and the equation of the directrix is x â€“ y â€“ 3 = 0 and e = 1/2.

Let P(x, y) by any point on the parabola

Squaring both sides, we have

â‡’ 8x

â‡’ 7x

Hence, the correct option is (a).

Solution.

Equation of the ellipse is

3x

â‡’

Here a

b

Length of the latus rectum =

Hence, the correct option is (d).

Q.56. If e is the eccentricity of the ellipse then

Solution.

Given equation is

âˆ´Eccentricity e

â‡’

Hence, the correct option is (b).

Solution.

Length of the latus rectum of the hyperbola

Distance between the foci = 2ae

Transverse axis = 2a

and Conjugate axis = 2b

âˆ´)

[from eq. (i)]

â‡’

Now b

â‡’ 4a = a

Hence, the correct option is (c).

Solution.

We know that the distance between the foci = 2ae

âˆ´ 2ae = 16 â‡’ ae = 8

Given that e = âˆš2

âˆ´ 2a = 8 â‡’ a = 4 âˆš2

Now b

â‡’ b

â‡’ b

So, the equation of the hyperbola is

Hence, the correct option is (a).

Solution.

Given that e = 3/2

and foci = (Â± ae, 0) = (Â± 2, 0)

âˆ´ ae = 2

Now we know that b

â‡’

So, the equation of the hyperbola is

Hence, the correct option is (a).

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233 docs

### Revision Notes - Parabola

- Doc | 2 pages
### Solved Examples - Parabola

- Doc | 2 pages

- Solved Examples - Hyperbola
- Doc | 2 pages
- Revision Notes - Hyperbola
- Doc | 3 pages
- Solved Examples - Ellipse
- Doc | 2 pages
- Revision Notes - Ellipse
- Doc | 2 pages