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All questions of Coordinate Geometry- Hyperbola for A Level Exam
The equation of the tangent lines to the hyperbola x2 – 2y2 = 18 which are perpendicular to the line y = x are- a)y = x ± 3
- b)y = –x ± 3
- c)2x + 3y + 4 = 0
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The equation of the tangent lines to the hyperbola x2 – 2y2 = 18 which are perpendicular to the line y = x are
a)
y = x ± 3
b)
y = –x ± 3
c)
2x + 3y + 4 = 0
d)
None of these
|
Preeti Khanna answered |
Equation of line perpendicular to x−y=0 is given by
y=−x+c
Also this line is tangent to the hyperbola x2−2y2=18
So we have m=−1, a2=18, b2=9
Thus Using condition of tangency c2 = a2m2−b2
= 18−9=9
⇒ c = ±3
Hence required equation of tangent is x+y = ±3
y=−x+c
Also this line is tangent to the hyperbola x2−2y2=18
So we have m=−1, a2=18, b2=9
Thus Using condition of tangency c2 = a2m2−b2
= 18−9=9
⇒ c = ±3
Hence required equation of tangent is x+y = ±3
Foot of normals drawn from the point p(h,k) to the hyperbola 
will always lie on the conic- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Foot of normals drawn from the point p(h,k) to the hyperbola will always lie on the conic
will always lie on the conica)
b)
c)
d)
|
Vijay Kumar answered |
Equation of normal at any point (x1, y1) is 
it passes through
P (h, k) then
thus (x1, y1) lie on the conic
it passes through P (h, k) then
thus (x1, y1) lie on the conic The asymptotes of the hyperbola xy–3x–2y = 0 are- a)x – 2 = 0 and y – 3 = 0
- b)x – 3 = 0 and y – 2 = 0
- c)x + 2 = 0 and y + 3 = 0
- d)x +3 = 0 and y + 2 = 0
Correct answer is option 'A'. Can you explain this answer?
The asymptotes of the hyperbola xy–3x–2y = 0 are
a)
x – 2 = 0 and y – 3 = 0
b)
x – 3 = 0 and y – 2 = 0
c)
x + 2 = 0 and y + 3 = 0
d)
x +3 = 0 and y + 2 = 0
|
Suresh Reddy answered |
xy - 3x - 2y + λ = 0.
Then abc + 2fgh − af2 − bg2 − ch2 = 0
⇒ 3/2 − λ/4 = 0
⇒ λ = 6
∴ Equation of asymptotes is xy-3x-2y+6=0
⇒ (x-2)(y-3)=0
⇒x - 2 = 0 and y - 3 = 0
Then abc + 2fgh − af2 − bg2 − ch2 = 0
⇒ 3/2 − λ/4 = 0
⇒ λ = 6
∴ Equation of asymptotes is xy-3x-2y+6=0
⇒ (x-2)(y-3)=0
⇒x - 2 = 0 and y - 3 = 0
Equation of the chord of the hyperbola 25x2 – 16y2 = 400 which is bisected at the point (6, 2) is- a)16x – 75y = 418
- b)75x – 16y = 418
- c)25x – 4y = 400
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Equation of the chord of the hyperbola 25x2 – 16y2 = 400 which is bisected at the point (6, 2) is
a)
16x – 75y = 418
b)
75x – 16y = 418
c)
25x – 4y = 400
d)
None of these
|
Hansa Sharma answered |
Given hyperbola is 25x2−16y2=400
If (6, 2) is the midpoint of the chord, then equation of chord is T = S1
⇒25(6x)−16(2y)=25(36)−16(4)
⇒75x−16y=450−32
⇒75x−16y=418
If (6, 2) is the midpoint of the chord, then equation of chord is T = S1
⇒25(6x)−16(2y)=25(36)−16(4)
⇒75x−16y=450−32
⇒75x−16y=418
The equation of the hyperbola whose foci are (6, 5), (–4, 5) and eccentricity 5/4 is- a)
- 
= 1 - b)
- 
= 1 - c)
- 
= - 1 - d)None of these
Correct answer is option 'A'. Can you explain this answer?
The equation of the hyperbola whose foci are (6, 5), (–4, 5) and eccentricity 5/4 is
a)
- = 1b)
- = 1c)
- = - 1d)
None of these
|
Om Desai answered |
Let the centre of hyperbola be (α,β)
As y=5 line has the foci, it also has the major axis.
∴ [(x−α)2]/a2 − [(y−β)2]/b2 = 1
Midpoint of foci = centre of hyperbola
∴ α=1,β=5
Given, e= 5/4
We know that foci is given by (α±ae,β)
∴ α+ae=6
⇒1+(5/4a)=6
⇒ a=4
Using b2 = a2(e2 − 1)
⇒ b2=16((25/16)−1)=9
∴ Equation of hyperbola ⇒ [(x−1)2]/16−[(y−5)2]/9=1
As y=5 line has the foci, it also has the major axis.
∴ [(x−α)2]/a2 − [(y−β)2]/b2 = 1
Midpoint of foci = centre of hyperbola
∴ α=1,β=5
Given, e= 5/4
We know that foci is given by (α±ae,β)
∴ α+ae=6
⇒1+(5/4a)=6
⇒ a=4
Using b2 = a2(e2 − 1)
⇒ b2=16((25/16)−1)=9
∴ Equation of hyperbola ⇒ [(x−1)2]/16−[(y−5)2]/9=1
The eccentricity of the hyperbola 4x2–9y2–8x=32 is- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The eccentricity of the hyperbola 4x2–9y2–8x=32 is
a)
b)
c)
d)
|
Top Rankers answered |
4x2−9y2−8x=32
⇒4(x2−2x)−9y2 = 32
⇒4(x2−2x+1)−9y2 = 32 + 4 = 36
⇒(x−1)2]/9 − [y2]/4 = 1
⇒a2=9, b2=4
∴e=[1+b2/a2]1/2
= [(13)1/2]/3
⇒4(x2−2x)−9y2 = 32
⇒4(x2−2x+1)−9y2 = 32 + 4 = 36
⇒(x−1)2]/9 − [y2]/4 = 1
⇒a2=9, b2=4
∴e=[1+b2/a2]1/2
= [(13)1/2]/3
Area of triangle formed by tangent to the hyperbola xy = 16 at (16, 1) and co-ordinate axes equals- a)8
- b)16
- c)32
- d)64
Correct answer is option 'C'. Can you explain this answer?
Area of triangle formed by tangent to the hyperbola xy = 16 at (16, 1) and co-ordinate axes equals
a)
8
b)
16
c)
32
d)
64
|
Riya Banerjee answered |
Differentiating xy=16, (xdy)/x+y=0
⇒ dy/dx=−y/x=−1/16= Slope of tangent
⇒ Its (tangent's) equation : y=−1=−1/16(x−16)
⇒ 16y−16=−x+16 ⇒ 16y=−x−32
⇒ 16y+x+32=0
It will cut x−axis at A(−32, 0)
& y−axis at B(0, −2)
⇒ Area of △OAB= 1/2×2×32
⇒ 32
⇒ dy/dx=−y/x=−1/16= Slope of tangent
⇒ Its (tangent's) equation : y=−1=−1/16(x−16)
⇒ 16y−16=−x+16 ⇒ 16y=−x−32
⇒ 16y+x+32=0
It will cut x−axis at A(−32, 0)
& y−axis at B(0, −2)
⇒ Area of △OAB= 1/2×2×32
⇒ 32
The equation of the transverse and conjugate axes of a hyperbola are respectively x + 2y – 3 = 0, 2x – y + 4 = 0 and their respective lengths are 
The equation of the hyperbola is- a)
- b)
- c)
- d)2(x + 2y -3)2 -3 (2x - y + 4)2 = 1
Correct answer is option 'A'. Can you explain this answer?
The equation of the transverse and conjugate axes of a hyperbola are respectively x + 2y – 3 = 0, 2x – y + 4 = 0 and their respective lengths are The equation of the hyperbola is
The equation of the hyperbola isa)
b)
c)
d)
2(x + 2y -3)2 -3 (2x - y + 4)2 = 1
|
|
Vikas Kapoor answered |
Equation of the hyperbola is
Where a1x + b1y + c1 = 0, b1x - a1y + c2 = 0 are conjugate and transverse axes respectively and a, b are lengths of semitransverse and semiconjugate axes respectively.
Where a1x + b1y + c1 = 0, b1x - a1y + c2 = 0 are conjugate and transverse axes respectively and a, b are lengths of semitransverse and semiconjugate axes respectively.
The locus of the point of intersection of the lines √3x - y - 4√3k = 0 and √3kx + ky - 4√3 = 0 for different values of k is- a)Ellipse
- b)Parabola
- c)Circle
- d)Hyperbola
Correct answer is option 'D'. Can you explain this answer?
The locus of the point of intersection of the lines √3x - y - 4√3k = 0 and √3kx + ky - 4√3 = 0 for different values of k is
a)
Ellipse
b)
Parabola
c)
Circle
d)
Hyperbola
|
Mohit Rajpoot answered |
Given equation of line are
√3x−y−4√3k=0 …(i)
and √3kx+ky−4√3=0
From Eq. (i) 4√3–√k=3–√x−y
⇒ k=(√3x−y)/4√3
put in Eq. (ii), we get
√3x(√3x−y)/4√3)+((√3x−y)/4√3)y−4√3=0
⇒1/4(√3x2−xy)+1/4(xy−y2/√3)−4√3=0
⇒√3/4x2−y2/4√3-4√3=0
⇒3x2−y2−48=0
⇒3x2−y2=48,which is hyperbola.
√3x−y−4√3k=0 …(i)
and √3kx+ky−4√3=0
From Eq. (i) 4√3–√k=3–√x−y
⇒ k=(√3x−y)/4√3
put in Eq. (ii), we get
√3x(√3x−y)/4√3)+((√3x−y)/4√3)y−4√3=0
⇒1/4(√3x2−xy)+1/4(xy−y2/√3)−4√3=0
⇒√3/4x2−y2/4√3-4√3=0
⇒3x2−y2−48=0
⇒3x2−y2=48,which is hyperbola.
such that DAOB (where `O' is the origin) is an equilateral triangle, then the eccentricity e of the hyperbola satisfies
If the latus rectum of an hyperbola be 8 and eccentricity be 3/√5 then the equation of the hyperbola is- a)4x2 – 5y2 = 100
- b)5x2 – 4y2 = 100
- c)4x2 + 5y2 = 100
- d)5x2 + 4y2 = 100
Correct answer is option 'A'. Can you explain this answer?
If the latus rectum of an hyperbola be 8 and eccentricity be 3/√5 then the equation of the hyperbola is
a)
4x2 – 5y2 = 100
b)
5x2 – 4y2 = 100
c)
4x2 + 5y2 = 100
d)
5x2 + 4y2 = 100
|
Neha Joshi answered |
Give eccentricity of the hyperbola is, e= 3/(5)1/2
⇒ (b2)/(a2) = 4/5..(1)
And latus rectum is =2(b2)/a=8
⇒ b2/a=4…(2)
By (2)/(1) a=5
∴ b2=20
Hence required hyperbola is, (x2)/25−(y2)/20=1
⇒ 4x2−5y2 =100
⇒ (b2)/(a2) = 4/5..(1)
And latus rectum is =2(b2)/a=8
⇒ b2/a=4…(2)
By (2)/(1) a=5
∴ b2=20
Hence required hyperbola is, (x2)/25−(y2)/20=1
⇒ 4x2−5y2 =100
If the product of the perpendicular distances from any point on the hyperbola 
of eccentricity e= on its asymptotes is equal to 6, then the length of the transverse axis of the hyperbola is- a)3
- b)6
- c)8
- d)12
Correct answer is option 'B'. Can you explain this answer?
If the product of the perpendicular distances from any point on the hyperbola of eccentricity e= on its asymptotes is equal to 6, then the length of the transverse axis of the hyperbola is
of eccentricity e= on its asymptotes is equal to 6, then the length of the transverse axis of the hyperbola isa)
3
b)
6
c)
8
d)
12
|
New Words answered |
e2 = (a2 / b2) + 1
3 = (a2 / b2) + 1
a2/b2 = 2
a2 = 2b2
Product of perpendicular distance of any point on hyperbola = (a2b2)/a2 + b2
= [2(b2).b2]/(2b2 + b2)
= [2b2]/3 = 6
= 2b2 = 18
=> b2 = 9
=> b = 3
Length (2b) = 2(3)
= 6
3 = (a2 / b2) + 1
a2/b2 = 2
a2 = 2b2
Product of perpendicular distance of any point on hyperbola = (a2b2)/a2 + b2
= [2(b2).b2]/(2b2 + b2)
= [2b2]/3 = 6
= 2b2 = 18
=> b2 = 9
=> b = 3
Length (2b) = 2(3)
= 6
The number of points on X-axis which are at a distance c units (c < 3) from (2, 3) is- a)1
- b)0
- c)3
- d)2
Correct answer is option 'B'. Can you explain this answer?
The number of points on X-axis which are at a distance c units (c < 3) from (2, 3) is
a)
1
b)
0
c)
3
d)
2
|
|
Mohit Rajpoot answered |
Distance of 'c' units from (2,3)
Let the no: of points be (x,0)
By distance formula
{(2−x)2+(3−0)2}=c
4−4x+x2+9=c
⇒x2−4x+13 = c:c=2,2
There are the points of c,such that when they are applied back to the equations,the number of points will become zero.
Let the no: of points be (x,0)
By distance formula
{(2−x)2+(3−0)2}=c
4−4x+x2+9=c
⇒x2−4x+13 = c:c=2,2
There are the points of c,such that when they are applied back to the equations,the number of points will become zero.
Locus of the middle points of the parallel chords with gradient m of the rectangular hyperbola xy = c2 is- a)y + mx = 0
- b)y – mx = 0
- c)my – mx = 0
- d)my + x = 0
Correct answer is option 'A'. Can you explain this answer?
Locus of the middle points of the parallel chords with gradient m of the rectangular hyperbola xy = c2 is
a)
y + mx = 0
b)
y – mx = 0
c)
my – mx = 0
d)
my + x = 0
|
|
Rohit Shah answered |
Let the mid point of chord be p (h,k) The equation of chord with p as mid point is Kx + hy =2hk The given slope m of the chord is m Hence -k/h =m Therefore required ans is y+mx =0 .
The equation of a line passing through the centre of a rectangular hyperbola is x – y –1 = 0. if one of its asymptote is 3x-4y-6 = 0, then equation of its other asymptote is- a)4x + 3y + 17 = 0
- b)4x + 3y -17 = 0
- c)4x + 3y -15 = 0
- d)4x + 3y + 15 = 0
Correct answer is option 'B'. Can you explain this answer?
The equation of a line passing through the centre of a rectangular hyperbola is x – y –1 = 0. if one of its asymptote is 3x-4y-6 = 0, then equation of its other asymptote is
a)
4x + 3y + 17 = 0
b)
4x + 3y -17 = 0
c)
4x + 3y -15 = 0
d)
4x + 3y + 15 = 0
|
Ishita Reddy answered |
Pt of intersection of x-y-1=0 and 3x-4y-6=0 is (-2,-3) other asymptote will be in the form of 4x + 3y + λ =0 and it should pass through (-2,-3). Thus λ = -17
A tangent to the parabola x2 = 4ay meets the hyperbola x2 - y2 = a2 at two points P and Q, then midpoint of P and Q lies on the curve- a)y3 = x (y- a)
- b)y3 = x2 (y- a)
- c)y2 = x2 (y- a)
- d)y2 = x3 (a- y)
Correct answer is option 'B'. Can you explain this answer?
A tangent to the parabola x2 = 4ay meets the hyperbola x2 - y2 = a2 at two points P and Q, then midpoint of P and Q lies on the curve
a)
y3 = x (y- a)
b)
y3 = x2 (y- a)
c)
y2 = x2 (y- a)
d)
y2 = x3 (a- y)
|
Nitya Basak answered |
Equation of tangent to parabola y = mx- am2 .......(1) equation of chord of hyperbola whose midpoint is (h, k) is hx - ky = h2 - k2 ...... (2) form (1) and (2)
The asymptotes of the curve 2x2 + 5xy + 2y2 + 4x + 5y = 0 are given by- a)2x2 + 5xy + 2y2 + 4x + 5y + 2 = 0
- b)2x2 + 5xy + 2y2 + 4x + 5y - 2 = 0
- c)2x2 + 5xy + 2y2 + 4x + 5y - 4 = 0
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The asymptotes of the curve 2x2 + 5xy + 2y2 + 4x + 5y = 0 are given by
a)
2x2 + 5xy + 2y2 + 4x + 5y + 2 = 0
b)
2x2 + 5xy + 2y2 + 4x + 5y - 2 = 0
c)
2x2 + 5xy + 2y2 + 4x + 5y - 4 = 0
d)
None of these
|
|
Baishali Tiwari answered |
The hyperbola is given by
2x2 + 5xy + 2y2 + 4x + 5y = 0 ............................(i)
Since the equation of hyperbola will differ from equation of asymptote by a constant. So, equation of asymptote is
2x2 + 5xy + 2y3 + 4x + 5y + λ = 0 ......................(ii)
If (ii) represents 2 straight lines, we must have
abc + 2fgh - af2- bg2 - ch2 = 0
or 22λ + 2.5 / 2.4 / 2.5 / 2 - 2.25 / 4 - 2.4 - λ (25 / 4) = 0
or 9λ = 18 ∴ λ = 2
2x2 + 5xy + 2y3 + 4x + 5y + 2 = 0
2x2 + 5xy + 2y2 + 4x + 5y = 0 ............................(i)
Since the equation of hyperbola will differ from equation of asymptote by a constant. So, equation of asymptote is
2x2 + 5xy + 2y3 + 4x + 5y + λ = 0 ......................(ii)
If (ii) represents 2 straight lines, we must have
abc + 2fgh - af2- bg2 - ch2 = 0
or 22λ + 2.5 / 2.4 / 2.5 / 2 - 2.25 / 4 - 2.4 - λ (25 / 4) = 0
or 9λ = 18 ∴ λ = 2
2x2 + 5xy + 2y3 + 4x + 5y + 2 = 0
t ∈ R represents
The equations x = at2, y = 4at ; t ∈ R represent- a)a parabola
- b)a circle
- c)an ellipse
- d)a hyperbola
Correct answer is option 'A'. Can you explain this answer?
The equations x = at2, y = 4at ; t ∈ R represent
a)
a parabola
b)
a circle
c)
an ellipse
d)
a hyperbola
|
Anjali Mehla answered |
Firstly I think you have to know the equation of a parabola and it's proff
The line y = c is a tangent to the parabola 7/2 if c is equal to- a)a
- b)0
- c)2a
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
The line y = c is a tangent to the parabola 7/2 if c is equal to
a)
a
b)
0
c)
2a
d)
none of these
|
|
Sravya Nair answered |
y = x is tangent to the parabola
y=ax2+c
if a= then c=?
y′ =2ax
y’ = 2(7/2)x =1
x = 1/7
1/7 = 2(1/7)2 + c
c = 1/7 * 2/49
c = 7/2
y=ax2+c
if a= then c=?
y′ =2ax
y’ = 2(7/2)x =1
x = 1/7
1/7 = 2(1/7)2 + c
c = 1/7 * 2/49
c = 7/2
A hyperbola passing through origin has 3x – 4y – 1 = 0 and 4x – 3y – 6 = 0 as its asymptotes. Then the equation of its transverse axis is- a)x – y – 5 = 0
- b)x + y + 1 = 0
- c)x + y – 5 = 0
- d)x – y – 1 = 0
Correct answer is option 'C'. Can you explain this answer?
A hyperbola passing through origin has 3x – 4y – 1 = 0 and 4x – 3y – 6 = 0 as its asymptotes. Then the equation of its transverse axis is
a)
x – y – 5 = 0
b)
x + y + 1 = 0
c)
x + y – 5 = 0
d)
x – y – 1 = 0
|
Hiral Rane answered |
Given Information:
- Equation of one asymptote: 3x - 4y - 1 = 0
- Equation of the other asymptote: 4x - 3y - 6 = 0
Explanation:
Finding the Slope of the Asymptotes:
1. Write the equations of the asymptotes in the form of y = mx + c.
2. Compare the coefficients of x and y to find the slopes of the asymptotes.
- For the first asymptote: slope = 3/4
- For the second asymptote: slope = 4/3
Equation of Transverse Axis:
3. Since the hyperbola passes through the origin, its center is also at the origin.
4. The transverse axis of the hyperbola passes through the center and is parallel to the asymptotes.
5. The equation of the transverse axis in the form of y = mx + c is given by y = x.
6. To find the equation in the form Ax + By + C = 0, we rearrange y = x to get x - y = 0.
7. Therefore, the equation of the transverse axis is x + y = 0, which can be rewritten as x + y - 0 = 0 or x + y = 0.
Conclusion:
The equation of the transverse axis of the hyperbola passing through the origin with asymptotes 3x - 4y - 1 = 0 and 4x - 3y - 6 = 0 is x + y = 0, which is option 'C'.
- Equation of one asymptote: 3x - 4y - 1 = 0
- Equation of the other asymptote: 4x - 3y - 6 = 0
Explanation:
Finding the Slope of the Asymptotes:
1. Write the equations of the asymptotes in the form of y = mx + c.
2. Compare the coefficients of x and y to find the slopes of the asymptotes.
- For the first asymptote: slope = 3/4
- For the second asymptote: slope = 4/3
Equation of Transverse Axis:
3. Since the hyperbola passes through the origin, its center is also at the origin.
4. The transverse axis of the hyperbola passes through the center and is parallel to the asymptotes.
5. The equation of the transverse axis in the form of y = mx + c is given by y = x.
6. To find the equation in the form Ax + By + C = 0, we rearrange y = x to get x - y = 0.
7. Therefore, the equation of the transverse axis is x + y = 0, which can be rewritten as x + y - 0 = 0 or x + y = 0.
Conclusion:
The equation of the transverse axis of the hyperbola passing through the origin with asymptotes 3x - 4y - 1 = 0 and 4x - 3y - 6 = 0 is x + y = 0, which is option 'C'.
A and B are two distinct points, Locus of a point P satisfying |PA| + |PB| = 2k, a constant is- a)nothing can be said
- b)the line segment [AB]
- c)an ellipse
- d)a hyperbola
Correct answer is option 'A'. Can you explain this answer?
A and B are two distinct points, Locus of a point P satisfying |PA| + |PB| = 2k, a constant is
a)
nothing can be said
b)
the line segment [AB]
c)
an ellipse
d)
a hyperbola
|
|
Ram Mohith answered |
I think the locus of P is an ellipse because if you take any point on the ellipse then the sum of distance from focii to that
The ellipse 4x2 + 9y2 = 36 and the hyperbola 4x2 – y2 = 4 have the same foci and they intersect at right angles then the equation of the circle through the points of intersection of two conics is- a)x2 + y2 = 5
- b)√5 (x2+y2) – 3x – 4y = 0
- c)√5 (x2+y2)+3x+4y=0
- d)x2 + y2 = 25
Correct answer is option 'A'. Can you explain this answer?
The ellipse 4x2 + 9y2 = 36 and the hyperbola 4x2 – y2 = 4 have the same foci and they intersect at right angles then the equation of the circle through the points of intersection of two conics is
a)
x2 + y2 = 5
b)
√5 (x2+y2) – 3x – 4y = 0
c)
√5 (x2+y2)+3x+4y=0
d)
x2 + y2 = 25
|
|
Rithika Desai answered |
The angle between the tangents drawn from the origin to the circle = (x−7)2+(y+1)2 = 25 is - a)π/8
- b)π/2
- c)π/6
- d)π/3
Correct answer is option 'B'. Can you explain this answer?
The angle between the tangents drawn from the origin to the circle = (x−7)2+(y+1)2 = 25 is
a)
π/8
b)
π/2
c)
π/6
d)
π/3
|
Shalini Yadav answered |
Let the equation of tangent drawn from (0,0) to the circle be y=mx. Then, p = a ⇒ 7m+1/(m2+1)1/2= 5
⇒24m2 + 14m−24=0
⇒12m2 + 7m−12=0
⇒m1m2 = −12/12 =−1
∴ Required angle = π/2
⇒24m2 + 14m−24=0
⇒12m2 + 7m−12=0
⇒m1m2 = −12/12 =−1
∴ Required angle = π/2
If the equation 4x2 + ky2 = 18 represents a rectangular hyperbola, then k =- a)4
- b)–4
- c)3
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
If the equation 4x2 + ky2 = 18 represents a rectangular hyperbola, then k =
a)
4
b)
–4
c)
3
d)
none of these
|
Tushar Choudhury answered |
To determine the value of k, we need to compare the given equation to the standard equation of a rectangular hyperbola, which is:
(x^2 / a^2) - (y^2 / b^2) = 1
In this equation, a represents the distance from the center to the vertices along the x-axis, and b represents the distance from the center to the vertices along the y-axis.
Comparing the given equation to the standard equation, we can see that a^2 = 4 and b^2 = 18. Therefore, a = 2 and b = √18.
Since the equation represents a rectangular hyperbola, a and b must be positive. So, b = √18 = 3√2.
Now, let's compare the values of a and b with the given options for k:
a) 4
b) 3√2
Since b = 3√2, the correct value of k is option b) 3√2.
(x^2 / a^2) - (y^2 / b^2) = 1
In this equation, a represents the distance from the center to the vertices along the x-axis, and b represents the distance from the center to the vertices along the y-axis.
Comparing the given equation to the standard equation, we can see that a^2 = 4 and b^2 = 18. Therefore, a = 2 and b = √18.
Since the equation represents a rectangular hyperbola, a and b must be positive. So, b = √18 = 3√2.
Now, let's compare the values of a and b with the given options for k:
a) 4
b) 3√2
Since b = 3√2, the correct value of k is option b) 3√2.
a = 2 and c2 = a2m2 - b2The locus of the point of intersection of tangents drawn at the extremities of normal chords to hyperbola xy = c2- a)(x2 - y2)2 + 4c2xy = 0
- b)(x2 + y2)2 + 4c2xy = 0
- c)(x2 - y2)2 + 4c xy = 0
- d)(x2 + y2)2 + 4cxy = 0
Correct answer is option 'A'. Can you explain this answer?
The locus of the point of intersection of tangents drawn at the extremities of normal chords to hyperbola xy = c2
a)
(x2 - y2)2 + 4c2xy = 0
b)
(x2 + y2)2 + 4c2xy = 0
c)
(x2 - y2)2 + 4c xy = 0
d)
(x2 + y2)2 + 4cxy = 0
|
Sinjini Datta answered |
Method to Solve :
The asymptotes of the hyperbola hx + ky = xy are- a)x - k = 0, y - h = 0
- b)x + h = 0, y + k = 0
- c)x - h = 0, y - k = 0
- d)x + k = 0, y + k = 0
Correct answer is option 'A'. Can you explain this answer?
The asymptotes of the hyperbola hx + ky = xy are
a)
x - k = 0, y - h = 0
b)
x + h = 0, y + k = 0
c)
x - h = 0, y - k = 0
d)
x + k = 0, y + k = 0
|
|
Rajdeep Iyer answered |
Asymptotes of a Hyperbola
Definition: The asymptotes of a hyperbola are the lines that the hyperbola approaches as the distance from the center of the hyperbola increases.
Asymptotes are important in the study of hyperbolas because they provide information about the behavior of the hyperbola as it extends towards infinity.
Equation of a Hyperbola
The general equation of a hyperbola is given by:
hx² - ky² = 1
where h and k are constants.
Finding the Asymptotes
To find the asymptotes of a hyperbola, we can follow these steps:
1. Rewrite the equation of the hyperbola in the standard form:
(x - h)²/a² - (y - k)²/b² = 1
where a and b are positive constants.
2. Compare the equation with the standard form:
(x - h)²/a² - (y - k)²/b² = 1
with the general form:
hx² - ky² = 1
By comparing the two equations, we can see that a² = 1/h and b² = 1/k.
3. Using the values of a and b, we can determine the slopes of the asymptotes. The slopes are given by:
m₁ = ±(b/a)
m₂ = ±(-b/a)
4. Finally, we can use the points (h, k) and the slopes to find the equations of the asymptotes using the point-slope form:
y - k = m₁(x - h)
y - k = m₂(x - h)
Answer Explanation
In the given equation hx - ky = xy, we can rewrite it in the standard form by rearranging the terms:
xy - hx - ky = 0
Comparing this equation with the standard form, we can see that a² = 1/h and b² = 1/k.
Since a and b are positive constants, the values of h and k must be positive.
Using the values of a and b, we can determine the slopes of the asymptotes:
m₁ = ±(b/a) = ±(1/k)/(1/h) = ±(h/k)
m₂ = ±(-b/a) = ±(-1/k)/(1/h) = ±(-h/k)
Using the point-slope form, the equations of the asymptotes are:
y - k = m₁(x - h) => y - k = ±(h/k)(x - h)
y - k = m₂(x - h) => y - k = ±(-h/k)(x - h)
Therefore, the correct answer is option 'A': x - k = 0, y - h = 0
Definition: The asymptotes of a hyperbola are the lines that the hyperbola approaches as the distance from the center of the hyperbola increases.
Asymptotes are important in the study of hyperbolas because they provide information about the behavior of the hyperbola as it extends towards infinity.
Equation of a Hyperbola
The general equation of a hyperbola is given by:
hx² - ky² = 1
where h and k are constants.
Finding the Asymptotes
To find the asymptotes of a hyperbola, we can follow these steps:
1. Rewrite the equation of the hyperbola in the standard form:
(x - h)²/a² - (y - k)²/b² = 1
where a and b are positive constants.
2. Compare the equation with the standard form:
(x - h)²/a² - (y - k)²/b² = 1
with the general form:
hx² - ky² = 1
By comparing the two equations, we can see that a² = 1/h and b² = 1/k.
3. Using the values of a and b, we can determine the slopes of the asymptotes. The slopes are given by:
m₁ = ±(b/a)
m₂ = ±(-b/a)
4. Finally, we can use the points (h, k) and the slopes to find the equations of the asymptotes using the point-slope form:
y - k = m₁(x - h)
y - k = m₂(x - h)
Answer Explanation
In the given equation hx - ky = xy, we can rewrite it in the standard form by rearranging the terms:
xy - hx - ky = 0
Comparing this equation with the standard form, we can see that a² = 1/h and b² = 1/k.
Since a and b are positive constants, the values of h and k must be positive.
Using the values of a and b, we can determine the slopes of the asymptotes:
m₁ = ±(b/a) = ±(1/k)/(1/h) = ±(h/k)
m₂ = ±(-b/a) = ±(-1/k)/(1/h) = ±(-h/k)
Using the point-slope form, the equations of the asymptotes are:
y - k = m₁(x - h) => y - k = ±(h/k)(x - h)
y - k = m₂(x - h) => y - k = ±(-h/k)(x - h)
Therefore, the correct answer is option 'A': x - k = 0, y - h = 0
A hyperbola has centre C and one focus at P(6, 8). If its two directrices are 3x + 4y + 10 = 0 and 3x + 4y -10 = 0 then CP =- a)14
- b)8
- c)10
- d)6
Correct answer is option 'C'. Can you explain this answer?
A hyperbola has centre C and one focus at P(6, 8). If its two directrices are 3x + 4y + 10 = 0 and 3x + 4y -10 = 0 then CP =
a)
14
b)
8
c)
10
d)
6
|
Rohit Yadav answered |
is nearest to 3x + 4y -10 = 0
CP = ae = 10
From any point on the hyperbola H1 : (x2/a2) - (y2/b2) = 1 tangents are drawn to the hyperbola .H2 : (x2/a2) - (y2/b2) = 2. The area cut-off by the chord of contact on the asymptotes of H2 is equal to- a)ab/2
- b)ab
- c)2 ab
- d)4 ab
Correct answer is option 'D'. Can you explain this answer?
From any point on the hyperbola H1 : (x2/a2) - (y2/b2) = 1 tangents are drawn to the hyperbola .H2 : (x2/a2) - (y2/b2) = 2. The area cut-off by the chord of contact on the asymptotes of H2 is equal to
a)
ab/2
b)
ab
c)
2 ab
d)
4 ab
|
Soumya Kumar answered |
Solution:
Let P(x1, y1) be any point on H1 and let the tangents drawn from P to the hyperbola H2 cut the asymptotes of H2 at points A and B. Let the area cut off by the chord of contact on the asymptotes of H2 be S.
1. Finding the equation of tangents to H2:
The equation of H2 is given by (x^2/a^2) - (y^2/b^2) = 2. Differentiating both sides w.r.t x, we get:
2x/a^2 - 2y/b^2(dy/dx) = 0
(dy/dx) = (x/a^2)(b^2/y)
The slope of the tangent at P is given by (b^2/y1)(x1/a^2). Therefore, the equation of the tangent passing through P is given by:
(y - y1) = (b^2/y1)(x1/a^2)(x - x1)
2. Finding the points of intersection of tangents with asymptotes:
The asymptotes of H2 are the lines y = (b/a)x and y = -(b/a)x. Substituting these values of y in the equation of the tangent and solving for x, we get the x-coordinates of the points of intersection of the tangent with the asymptotes. Substituting these values of x in the equation of the asymptotes, we get the corresponding y-coordinates.
3. Finding the area cut off by the chord of contact on the asymptotes:
The area cut off by the chord of contact on the asymptotes is the area of the trapezium formed by joining the points of intersection of the tangents with the asymptotes. Let A1, A2 be the points of intersection of the tangent passing through P with the asymptotes and let B1, B2 be the corresponding points for the other tangent. Then, the area cut off by the chord of contact is given by:
S = (1/2)(A1B1 + A2B2)(b/a)
4. Simplifying the expression for S:
We can simplify the expression for S by using the fact that the product of the distances of any point on H1 from the points (±a, 0) is equal to b^2. Therefore, we have:
x1^2 - a^2 = (b^2/a^2)y1^2
x1^2/a^2 - y1^2/b^2 = 1
Substituting the equation of H1 in the equation of the tangent to H2, we get:
(y - y1) = (b^2/y1)(x1/a^2)(x - x1)
y = (b^2/y1)(x1/a^2)(x - x1) + y1
Substituting this value of y in the equation of H1, we get a quadratic equation in x. Solving for x, we get:
x = ±(a^2/b^2)(y1^2 - b^2)
Substituting these values of x in the equation of the asymptotes, we get the corresponding y-coordinates. Therefore, we have:
A1B1 = 2a^2/b
A2B2 = 2a^
Let P(x1, y1) be any point on H1 and let the tangents drawn from P to the hyperbola H2 cut the asymptotes of H2 at points A and B. Let the area cut off by the chord of contact on the asymptotes of H2 be S.
1. Finding the equation of tangents to H2:
The equation of H2 is given by (x^2/a^2) - (y^2/b^2) = 2. Differentiating both sides w.r.t x, we get:
2x/a^2 - 2y/b^2(dy/dx) = 0
(dy/dx) = (x/a^2)(b^2/y)
The slope of the tangent at P is given by (b^2/y1)(x1/a^2). Therefore, the equation of the tangent passing through P is given by:
(y - y1) = (b^2/y1)(x1/a^2)(x - x1)
2. Finding the points of intersection of tangents with asymptotes:
The asymptotes of H2 are the lines y = (b/a)x and y = -(b/a)x. Substituting these values of y in the equation of the tangent and solving for x, we get the x-coordinates of the points of intersection of the tangent with the asymptotes. Substituting these values of x in the equation of the asymptotes, we get the corresponding y-coordinates.
3. Finding the area cut off by the chord of contact on the asymptotes:
The area cut off by the chord of contact on the asymptotes is the area of the trapezium formed by joining the points of intersection of the tangents with the asymptotes. Let A1, A2 be the points of intersection of the tangent passing through P with the asymptotes and let B1, B2 be the corresponding points for the other tangent. Then, the area cut off by the chord of contact is given by:
S = (1/2)(A1B1 + A2B2)(b/a)
4. Simplifying the expression for S:
We can simplify the expression for S by using the fact that the product of the distances of any point on H1 from the points (±a, 0) is equal to b^2. Therefore, we have:
x1^2 - a^2 = (b^2/a^2)y1^2
x1^2/a^2 - y1^2/b^2 = 1
Substituting the equation of H1 in the equation of the tangent to H2, we get:
(y - y1) = (b^2/y1)(x1/a^2)(x - x1)
y = (b^2/y1)(x1/a^2)(x - x1) + y1
Substituting this value of y in the equation of H1, we get a quadratic equation in x. Solving for x, we get:
x = ±(a^2/b^2)(y1^2 - b^2)
Substituting these values of x in the equation of the asymptotes, we get the corresponding y-coordinates. Therefore, we have:
A1B1 = 2a^2/b
A2B2 = 2a^
The eccentricity of the conic 9x2 − 16y2 = 144 is- a)4/3
- b)5/4
- c)√7
- d)4/5
Correct answer is option 'B'. Can you explain this answer?
The eccentricity of the conic 9x2 − 16y2 = 144 is
a)
4/3
b)
5/4
c)
√7
d)
4/5
|
|
Mihir Chaudhary answered |
+16y2=144 is:
To find the eccentricity of a conic, we need to first identify the type of conic. We can rewrite the given equation as:
9x^2/144 + 16y^2/144 = 1
Dividing both sides by 144, we get:
x^2/16 + y^2/9 = 1
This is the equation of an ellipse. To find the eccentricity, we need to first find the distance between the center of the ellipse and one of its foci. The center of the ellipse is at the point (0,0). We can find the length of the semi-major axis (a) and the semi-minor axis (b) using the equation:
a^2 = 16, b^2 = 9
a = 4, b = 3
The distance between the center and one of the foci (c) can be found using the equation:
c^2 = a^2 - b^2
c^2 = 16 - 9 = 7
c = sqrt(7)
The eccentricity (e) of the ellipse is given by the equation:
e = c/a
e = sqrt(7)/4
Therefore, the eccentricity of the conic 9x^2+16y^2=144 is sqrt(7)/4.
To find the eccentricity of a conic, we need to first identify the type of conic. We can rewrite the given equation as:
9x^2/144 + 16y^2/144 = 1
Dividing both sides by 144, we get:
x^2/16 + y^2/9 = 1
This is the equation of an ellipse. To find the eccentricity, we need to first find the distance between the center of the ellipse and one of its foci. The center of the ellipse is at the point (0,0). We can find the length of the semi-major axis (a) and the semi-minor axis (b) using the equation:
a^2 = 16, b^2 = 9
a = 4, b = 3
The distance between the center and one of the foci (c) can be found using the equation:
c^2 = a^2 - b^2
c^2 = 16 - 9 = 7
c = sqrt(7)
The eccentricity (e) of the ellipse is given by the equation:
e = c/a
e = sqrt(7)/4
Therefore, the eccentricity of the conic 9x^2+16y^2=144 is sqrt(7)/4.
If x = 9 is the chord of contact of the hyperbola x2 - y2 = 9, then the equation of the tangent at one of the points of contact is - a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
If x = 9 is the chord of contact of the hyperbola x2 - y2 = 9, then the equation of the tangent at one of the points of contact is
a)
b)
c)
d)
|
Nilesh Goyal answered |
Solve x = 9 and x2 – y2 = 9 equation of tangent at point of contact is S1 = 0
The number of tangents and normals to the hyperbola 
of the slope 1 is- a)2, 2
- b)2, 0
- c)0, 0
- d)0, 2
Correct answer is option 'C'. Can you explain this answer?
The number of tangents and normals to the hyperbola of the slope 1 is
of the slope 1 isa)
2, 2
b)
2, 0
c)
0, 0
d)
0, 2
|
Pranavi Chavan answered |
Equation of tangents and normal with slope 'm' will be
A hyperbola, having the transverse axis of length 2 sin θ, is confocal with the ellipse 3x2 + 4y2 = 12. Then its equation is:- a)x2 cosec2θ - y2 sec2 θ = 1
- b)x2 sec2 θ - y2 cosec2θ = 1
- c)x2 sin2θ - y2 cos2θ = 1
- d)x2 cos2θ - y2 sin2θ = 1
Correct answer is option 'A'. Can you explain this answer?
A hyperbola, having the transverse axis of length 2 sin θ, is confocal with the ellipse 3x2 + 4y2 = 12. Then its equation is:
a)
x2 cosec2θ - y2 sec2 θ = 1
b)
x2 sec2 θ - y2 cosec2θ = 1
c)
x2 sin2θ - y2 cos2θ = 1
d)
x2 cos2θ - y2 sin2θ = 1
|
|
Ishita Reddy answered |
The given ellipse is 
Hence, the eccentricity e1, of the hyperbola is given by
1 = e1 sinθ ⇒ e1 = cosec θ ⇒ b2 = sin2θ (cosec2θ - 1) = cos2θ
Hence, the eccentricity e1, of the hyperbola is given by
1 = e1 sinθ ⇒ e1 = cosec θ ⇒ b2 = sin2θ (cosec2θ - 1) = cos2θ
Tangents are drawn to 3x2 - 2y2 = 6 from a point P. If these tangents intersects the coordinate axes at concyclic points, The locus of P is- a)x2 + y2 = 5
- b)
- c)x2 - y2 = 13
- d)
Correct answer is option 'C'. Can you explain this answer?
Tangents are drawn to 3x2 - 2y2 = 6 from a point P. If these tangents intersects the coordinate axes at concyclic points, The locus of P is
a)
x2 + y2 = 5
b)
c)
x2 - y2 = 13
d)
|
|
Mahi Reddy answered |
x2/2 - y2/3 = 1
= [y2 + (3)2]/[x2 - (2)2] = 1
=> [y2 + 9]/[x2 - 4] = 1
=> [y2 + 9] = [x2 - 4]
x2 - y2 = 13
= [y2 + (3)2]/[x2 - (2)2] = 1
=> [y2 + 9]/[x2 - 4] = 1
=> [y2 + 9] = [x2 - 4]
x2 - y2 = 13
Let a and b be non–zero real numbers. Then, the equation (ax2 + by2 + (C) (x2 – 5xy + 6y2) = 0 represents - a)four straight lines, when c = 0 and a, b are of the same sign
- b)two straight lines and a circle, when a = b, and c is of sign opposite to that of a
- c)two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a
- d)a circle and an ellipse, when a and b are of the same sign and c is of sign opposite to that of a.
Correct answer is option 'B'. Can you explain this answer?
Let a and b be non–zero real numbers. Then, the equation (ax2 + by2 + (C) (x2 – 5xy + 6y2) = 0 represents
a)
four straight lines, when c = 0 and a, b are of the same sign
b)
two straight lines and a circle, when a = b, and c is of sign opposite to that of a
c)
two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a
d)
a circle and an ellipse, when a and b are of the same sign and c is of sign opposite to that of a.
|
Bhavana Das answered |
x2 – 5xy + 6y2 = 0 represent a pair of lines passing through origin
ax2 + ay2 = - c
⇒ ax2 + ay2 + c = 0 represent a circle
ax2 + ay2 = - c
⇒ ax2 + ay2 + c = 0 represent a circle
The vertex of the parabola y2 = 4a(x−a) is- a)(0, 0)
- b)(a, 0)
- c)(0, a)
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The vertex of the parabola y2 = 4a(x−a) is
a)
(0, 0)
b)
(a, 0)
c)
(0, a)
d)
none of these
|
|
Harshad Das answered |
The vertex of the parabola y^2 = 4a(x) is (0,0).
If the tangent at the point (h, k) on the hyperbola 
meets the auxiallary circle of the hyperbola in two points whose ordinates y1, y2 then 
- a)1/k
- b)
- c)
- d)2/k
Correct answer is option 'D'. Can you explain this answer?
If the tangent at the point (h, k) on the hyperbola meets the auxiallary circle of the hyperbola in two points whose ordinates y1, y2 then
meets the auxiallary circle of the hyperbola in two points whose ordinates y1, y2 then a)
1/k
b)
c)
d)
2/k
|
|
Bibek Khanna answered |
Equation of the tangent at (h, k) is
Equation of auxiallary circle x2 + y2 = a2
Equation of auxiallary circle x2 + y2 = a2
and 
are
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