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JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 211-240 of 429 questions

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211 Easy
211 of 429
Application +4 | -1
If the normal at 'θ\theta' on the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 meets the transverse axis at GG, and AA and A′A^{\prime} are the vertices of the hyperbola, then AG⋅A′G=A G \cdot A^{\prime} G=
View answer and explanation
Correct option: B
a2(e4sec⁡2θ−1)a^2\left(e^4 \sec ^2 \theta-1\right)
Option B (a2(e4sec⁡2θ−1)a^2\left(e^4 \sec ^2 \theta-1\right)) is correct because it represents the product of the distances AGAG and A′GA'G, calculated using the coordinates of the vertices and the point where the normal intersects the transverse axis.
212 Medium
212 of 429
Application +4 | -1
If e1{e_1} and e2{e_2} are the eccentricities of a hyperbola 3x2−3y2=253{x^2} - 3{y^2} = 25 and its conjugate, then
View answer and explanation
Correct option: B
e12+e22=4e_1^2 + e_2^2 = 4
Option B (e12+e22=4e_1^2 + e_2^2 = 4) is correct because the eccentricities of the hyperbola and its conjugate are both 2\sqrt{2}, and the sum of their squares is 4.
213 Easy
213 of 429
Application +4 | -1
The locus of the middle points of chords of hyperbola 3x2−2y2+4x−6y=03 x^2-2 y^2+4 x-6 y=0 parallel to y=2xy=2 x is
View answer and explanation
Correct option: A
3x−4y=43 x-4 y=4
Option A (3x−4y=43 x-4 y=4) is correct because the locus of the midpoints of chords parallel to a given line can be found using the T = S1 method, and then substituting the slope condition.
214 Medium
214 of 429
Application +4 | -1
If the normals drawn at four concyclic points on a rectangular hyperbola xy=c2x y=c^2 meet at point (α,β)(\alpha, \beta) then the centre of the circle has the coordinates
View answer and explanation
Correct option: C
(α2,β2)\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)
Option C ((α2,β2)\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)) is correct because the center of the circle is the midpoint of the line joining the point of concurrency of normals and the center of the hyperbola.
215 Medium
215 of 429
Application +4 | -1
The product of lengths of perpendicular from any point on the hyperbola x2−y2=8{x^2} - {y^2} = 8 to its asymptotes is
View answer and explanation
Correct option: D
4
Option D (4) is correct because it represents the product of the lengths of the perpendiculars from any point on the given hyperbola to its asymptotes.
216 Easy
216 of 429
Application +4 | -1
A common tangent to the conics x2=6yx^{2}=6 y and 2x2−4y2=92 x^{2}-4 y^{2}=9 is:
View answer and explanation
Correct option: A
x−y=32x-y=\frac{3}{2}
Option A (x−y=32x-y=\frac{3}{2}) is correct because it represents a common tangent to both the given conics.
217 Easy
217 of 429
Application +4 | -1
The foci of a hyperbola coincide with the foci of the ellipse x225+y29=1\frac{x^2}{25}+\frac{y^2}{9}=1. Find the equation of the hyperbola, if its eccentricity is 2, is
View answer and explanation
Correct option: A
x24−y212=1\frac{x^2}{4}-\frac{y^2}{12}=1
Option A (x24−y212=1\frac{x^2}{4}-\frac{y^2}{12}=1) is correct because it correctly calculates the foci of the ellipse and uses them to find the equation of the hyperbola with the given eccentricity.
218 Medium
218 of 429
Analyzing +4 | -1
Let the major axis of a standard ellipse equals the transverse axis of a standard hyperbola and their director circles have radius equal to 2R2 R and RR respectively. If e1e_1 and e2e_2 are the eccentricities of the ellipse and hyperbola, then the correct relation is
View answer and explanation
Correct option: C
4e22−e12=64 e_2^2-e_1^2=6
Option C (4e22−e12=64 e_2^2-e_1^2=6) is correct because it is derived by relating the eccentricities of the ellipse and hyperbola to the radii of their director circles.
219 Easy
219 of 429
Application +4 | -1
If the tangent and normal to a rectangular hyperbola cut off intercepts a1a_1 and a2a_2 on one axis and b1b_1 and b2b_2 on the other, then
View answer and explanation
Correct option: A
a1a2+b1b2=0a_1 a_2+b_1 b_2=0
Option A (a1a2+b1b2=0a_1 a_2+b_1 b_2=0) is correct because the product of the intercepts of the tangent and normal on one axis plus the product of the intercepts on the other axis equals zero for a rectangular hyperbola.
220 Medium
220 of 429
Analyzing +4 | -1
If a circles x2+y2=a2x^2+y^2=a^2 and the rectangular hyperbola xy=c2x y=c^2 intersect in four points, (ctr,ctr)\left(c t_r, \frac{c}{t_r}\right), r=1,2,3,4r=1,2,3,4 then t1t2t3t4t_1 t_2 t_3 t_4 is equal to
View answer and explanation
Correct option: B
1
Option B is correct. Substituting the parametric equations of the hyperbola into the equation of the circle and solving for the product of the roots yields 1.
221 Easy
221 of 429
Application +4 | -1
If the latus rectum of a hyperbola be 8 and its eccentricity be 3/53 / \sqrt{5}, then the equation of the hyperbola is
View answer and explanation
Correct option: A
4x2−5y2=1004 x^2-5 y^2=100
Option A (4x2−5y2=1004 x^2-5 y^2=100) is correct because it satisfies the given conditions for the latus rectum and eccentricity of the hyperbola.
222 Easy
222 of 429
Analyzing +4 | -1
The one which does not represent a hyperbola is
View answer and explanation
Correct option: D
x2−y2=0{x^2} - {y^2} = 0
Option D (x2−y2=0{x^2} - {y^2} = 0) is correct because it represents a pair of straight lines, not a hyperbola.
223 Easy
223 of 429
Application +4 | -1
The mid-point of the chord intercepted by the hyperbola 9x2−16y2=1449{x^2} - 16{y^2} = 144 on the line 9x−8y−10=0,9x - 8y - 10 = 0, is
View answer and explanation
Correct option: D
(2,1)\left( {2,1} \right)
Option D ((2,1)) is correct because it is found by using the T=S' method to find the midpoint of the chord intercepted by the line and hyperbola.
224 Easy
224 of 429
Application +4 | -1
If the eccentricities of the hyperbolas x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 and y2b2−x2a2=1\frac{y^2}{b^2}-\frac{x^2}{a^2}=1 be ee and e1e_1, then find the value of 1e2+1e12\frac{1}{e^2}+\frac{1}{e_1^2}.
View answer and explanation
Correct option: A
1
Option A (1) is correct because the sum of the reciprocals of the squares of the eccentricities of the two hyperbolas is equal to 1.
225 Medium
225 of 429
Application +4 | -1
The straight line x+y=2px + y = \sqrt 2 p will touch the hyperbola 4x2−9y2=36,4{x^2} - 9{y^2} = 36, if
View answer and explanation
Correct option: D
2p2=52{p^2} = 5
Option D (2p2=52{p^2} = 5) is correct because the condition for the line to touch the hyperbola leads to the equation 2p2=52{p^2} = 5.
226 Easy
226 of 429
Application +4 | -1
If 2x−y+1=02 x-y+1=0 is a tangent to hyperbola x2a2+y216=1\frac{x^2}{a^2}+\frac{y^2}{16}=1, then which of the following are sides of a right angled triangle?
View answer and explanation
Correct option: A
2a,4,12 a, 4,1
Option A (2a,4,12a, 4, 1) is correct because the tangent condition to the hyperbola leads to a relationship between aa and the constant term, resulting in these side lengths forming a right-angled triangle.
227 Easy
227 of 429
Application +4 | -1
The equation of the hyperbola in the standard form (with transverse axis along the xx-axis) having the length of the latusrectum =9= 9 unit and eccentricity =54= \frac{5}{4}, is
View answer and explanation
Correct option: C
x264−y236=1\frac{{{x^2}}}{{64}} - \frac{{{y^2}}}{{36}} = 1
Option C (x264−y236=1\frac{{{x^2}}}{{64}} - \frac{{{y^2}}}{{36}} = 1) is correct because it satisfies the given latus rectum length and eccentricity for a hyperbola.
228 Easy
228 of 429
Application +4 | -1
Let a>0,b>0a>0, b>0. Let ee and ll respectively be the eccentricity and length of the latus rectum of the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. Let e′e^{\prime} and l′l^{\prime} respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If e2=1114l\mathrm{e}^2=\frac{11}{14} l and (e′)2=118l′\left(\mathrm{e}^{\prime}\right)^2=\frac{11}{8} l^{\prime}, then the value of 77a+44b77 a+44 b is equal to :
View answer and explanation
Correct option: D
130
Option D (130) is correct because it is the result of solving the equations derived from the properties of the hyperbola and its conjugate.
229 Easy
229 of 429
Application +4 | -1
If ee is the eccentricity of the hyperbola x218−y26=1\frac{x^2}{18}-\frac{y^2}{6}=1 and rr is the radius of the circle x2+y2−6x−18y+87=0,x^2 + y^2 - 6x - 18y + 87 = 0, then the value of erer is equal to:
View answer and explanation
Correct option: C
2
Option C (2) is correct because it involves calculating the eccentricity of the hyperbola and the radius of the circle, then multiplying them.
230 Easy
230 of 429
Application +4 | -1
The point on the curve 3x2−4y2=723{x^2} - 4{y^2} = 72, which is nearest to the line 3x+2y−1=03x + 2y - 1 = 0, is
View answer and explanation
Correct option: B
(6,−3)\left( {6, - 3} \right)
Option B ((6,−3)\left( {6, - 3} \right)) is correct because it's the point on the curve closest to the line, determined by minimizing the distance between the point and the line.
231 Easy
231 of 429
Application +4 | -1
Foci of the rectangular hyperbola are (±7,0\pm 7,0), the equation of the hyperbola is
View answer and explanation
Correct option: C
2x2−2y2=492 x^2-2 y^2=49
Option C (2x2−2y2=492 x^2-2 y^2=49) is correct because it represents a rectangular hyperbola with the given foci.
232 Medium
232 of 429
Application +4 | -1
The locus of the middle points of the portions of the tangents of the hyperbola. x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 included between the axes is
View answer and explanation
Correct option: C
b2x2−a2y2+4x2y2=0b^2 x^2-a^2 y^2+4 x^2 y^2=0
Option B (b2x2−a2y2+4x2y2=0b^2 x^2-a^2 y^2+4 x^2 y^2=0) is correct because finding the intercepts, midpoint, and eliminating θ\theta leads to this equation.
233 Easy
233 of 429
Application +4 | -1
Number of points from where perpendicular tangents to the curve x216−y225=1\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{{25}} = 1 can be drawn, is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the director circle of the given hyperbola is an imaginary circle, indicating that there are no points from which perpendicular tangents can be drawn.
234 Easy
234 of 429
Remembering +4 | -1
The reciprocal of the eccentricity of rectangular hyperbola is
View answer and explanation
Correct option: C
2\sqrt{2}
Option D (1/21 / \sqrt{2}) is incorrect. The eccentricity of a rectangular hyperbola is 2\sqrt{2}, so the reciprocal is 1/21/\sqrt{2}.
235 Easy
235 of 429
Application +4 | -1
If e1,e2e_1, e_2 are the eccentricities of the hyperbola 2x2−2y2=12 x^2-2 y^2=1 and the ellipse x2+2y2=2x^2+2 y^2=2 respectively. Then
View answer and explanation
Correct option: B
e1e2=1e_1 e_2=1
Option B (e1e2=1e_1 e_2=1) is correct because when calculating the eccentricities of the given hyperbola and ellipse, their product equals 1.
236 Easy
236 of 429
Application +4 | -1
The length of the transverse axis of a hyperbola is 7 and it passes through the point (5,−2)(5,-2). The equation of the hyperbola is
View answer and explanation
Correct option: C
449x2−51196y2=1\frac{4}{49} x^2-\frac{51}{196} y^2=1
Option C (449x2−51196y2=1\frac{4}{49} x^2-\frac{51}{196} y^2=1) is correct because it represents a hyperbola with a transverse axis of 7 and passes through the point (5,-2).
237 Easy
237 of 429
Application +4 | -1
If the normal at PP on the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 meets the transverse axis at G,SG, S is a foci and ee the eccentricity of the hyperbola then SG:SPS G: S P is equal to
View answer and explanation
Correct option: C
ee
Option C (ee) is correct because the ratio SG:SP is equal to the eccentricity of the hyperbola.
238 Medium
238 of 429
Application +4 | -1
The locus of middle points of chords of hyperbola 3x2−2y2+4x−6y=03{x^2} - 2{y^2} + 4x - 6y = 0 parallel to y=2xy = 2x is
View answer and explanation
Correct option: A
3x−4y=43x - 4y = 4
Option A (3x−4y=43x - 4y = 4) is correct because it represents the locus of the midpoints of chords of the hyperbola parallel to the given line, derived using the T = S1 method.
239 Hard
239 of 429
Application +4 | -1
The product of the perpendicular from two foci on any tangent to the hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1, is
View answer and explanation
Correct option: B
b2{b^2}
Option B (b2) is correct because the product of the perpendiculars from the foci to any tangent of the hyperbola is equal to the square of the semi-minor axis.
240 Easy
240 of 429
Remembering +4 | -1
The equation ax2+2hxy+by2+2gx+2fy+c=0a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 represents a rectangular hyperbola if
View answer and explanation
Correct option: A
Δ≠0,h2>ab,a+b=0\Delta \neq 0, h^2 > a b, a+b=0
Option A (Δ≠0,h2>ab,a+b=0\Delta \neq 0, h^2 > a b, a+b=0) is correct because these are the conditions for the general equation to represent a rectangular hyperbola.
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