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JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Easy

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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 271-298 of 298 questions
271 Easy
271 of 298
Application +4 | -1
Equation of the chord of the hyperbola 25x2−16y2=40025{x^2} - 16{y^2} = 400 which is bisected at the point (6,2)\left( {6,2} \right), is
View answer and explanation
Correct option: B
75x−16y=41875x - 16y = 418
Option B (75x−16y=41875x - 16y = 418) is correct because the equation of the chord bisected at a given point can be found using the formula T=S1T = S_1.
272 Easy
272 of 298
Application +4 | -1
If the eccentricity of a hyperbola is 3\sqrt 3, then the eccentricity of its conjugate hyperbola is
View answer and explanation
Correct option: C
32\sqrt {\frac{3}{2}}
Option C (32\sqrt{\frac{3}{2}}) is correct because it is calculated using the relationship between the eccentricities of a hyperbola and its conjugate hyperbola: 1e2+1e′2=1\frac{1}{e^2} + \frac{1}{e'^2} = 1.
273 Easy
273 of 298
Application +4 | -1
The locus of the foot of the perpendicular drawn from the origin to any tangent to the hyperbola x236−y216=1\frac{x^2}{36}-\frac{y^2}{16}=1 is
View answer and explanation
Correct option: A
(x2+y2)2=36x2−16y2\left(x^2+y^2\right)^2=36 x^2-16 y^2
Option A ((x2+y2)2=36x2−16y2(x^2+y^2)^2=36 x^2-16 y^2) is correct because eliminating 'm' from the tangent and perpendicular line equations gives this locus.
274 Easy
274 of 298
Application +4 | -1
e1,e2e_1, e_2 are respectively the eccentricities of the hyperbola x2−y2cosec⁡2θ=5x^2-y^2 \operatorname{cosec}^2 \theta=5 and the ellipse x2cosec⁡2θ+y2=5x^2 \operatorname{cosec}^2 \theta+y^2=5. If 0<θ<π/20 < \theta < \pi / 2 and e1=7e2e_1=\sqrt{7} e_2, then θ\theta is equal to
View answer and explanation
Correct option: C
π3\frac{\pi}{3}
Option C (π3\frac{\pi}{3}) is correct because by finding the eccentricities e1e_1 and e2e_2 in terms of θ\theta and using the given relation e1=7e2e_1 = \sqrt{7}e_2, we can solve for θ\theta.
275 Easy
275 of 298
Application +4 | -1
A hyperbola whose transverse axis is along the major axis of the conic x23+y24=4\frac{x^2}{3}+\frac{y^2}{4}=4 and has vertices at the foci of the conic. If the eccentricity of the hyperbola is 3/23 / 2, then which of the following points does not lie on it?
View answer and explanation
Correct option: B
(5,23)(5,2 \sqrt{3})
Option B (5,23)(5,2 \sqrt{{3}}) is correct because when substituted into the hyperbola equation, it does not satisfy the equation, indicating it does not lie on the hyperbola.
276 Easy
276 of 298
Understanding +4 | -1
The difference in focal distances of any point on the hyperbola x216−y29=1\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1 is
View answer and explanation
Correct option: A
8
Option A (8) is correct because the difference in focal distances of any point on a hyperbola is equal to the length of its transverse axis (2a).
277 Easy
277 of 298
Understanding +4 | -1
A point on the curve x2A2−y2B2=1\frac{x^2}{A^2}-\frac{y^2}{B^2}=1 is
View answer and explanation
Correct option: B
(Asec⁡θ,Btan⁡θ)(A \sec \theta, B \tan \theta)
Option B ((Asec⁡θ,Btan⁡θ)(A \sec \theta, B \tan \theta)) is correct because substituting x=Asec⁡θx = A \sec \theta and y=Btan⁡θy = B \tan \theta into the equation x2A2−y2B2=1\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1 satisfies the equation.
278 Easy
278 of 298
Application +4 | -1
The equation of the tangent to the hyperbola 4y2=x2−14{y^2} = {x^2} - 1 at the point (1,0), is
View answer and explanation
Correct option: A
x=1x = 1
Option A (x=1x = 1) is the correct answer because substituting the point (1,0) into the equation of the tangent yields x=1x = 1.
279 Easy
279 of 298
Application +4 | -1
If in a hyperbola, the distance between the foci is 10 and the transverse axis has length 8, then the length of its latusrectum is
View answer and explanation
Correct option: B
92\frac{9}{2}
Option B (92\frac{9}{2}) is correct because using the given information to find 'a' and 'b' and then applying the latus rectum formula results in 9/2.
280 Easy
280 of 298
Application +4 | -1
If the chords of contact of tangents from two points (x1,y1)\left(x_1, y_1\right) and (x2,y2)\left(x_2, y_2\right) to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 are at right angles, then x1x2y1y2\frac{x_1 x_2}{y_1 y_2} is equal to
View answer and explanation
Correct option: D
−a4b4-\frac{a^4}{b^4}
Option D (−a4b4-\frac{a^4}{b^4}) is correct because the product of the slopes of the chords of contact is -1, leading to the given ratio.
281 Easy
281 of 298
Analyzing +4 | -1
The line 2x+y=12 x+y=1 is tangent to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. If this line passes through the point of intersection of the nearest directrix and the xx-axis, then the eccentricity of the hyperbola is
View answer and explanation
Correct option: B
2
Option B (2) is correct because by using the tangent condition and the fact that the line passes through the directrix intersection with the x-axis, we can solve for the eccentricity.
282 Easy
282 of 298
Understanding +4 | -1
If ee and e1e_1 are the eccentricities of a hyperbola and its conjugate, then 1e2+1e12\frac{1}{e^2}+\frac{1}{e_1^2} is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because 1e2+1e12=1\frac{1}{e^2} + \frac{1}{e_1^2} = 1 for a hyperbola and its conjugate.
283 Easy
283 of 298
Application +4 | -1
The tangents to the hyperbola x2−y2=3{x^2} - {y^2} = 3 are parallel to the straight line 2x+y+8=02x + y + 8 = 0 at the following points:
View answer and explanation
Correct option: B
(2,−1),(−2,1)\left( {2, - 1} \right),\left( { - 2,1} \right)
Option B ((2,−1),(−2,1)\left( {2, - 1} \right),\left( { - 2,1} \right)) is correct because these are the points where tangents parallel to the given line touch the hyperbola, calculated using the tangent's slope.
284 Easy
284 of 298
Application +4 | -1
If H(x,y)=0H(x, y)=0 represents the equation of a hyperbola and A(x,y)=0,C(x,y)=0A(x, y)=0, C(x, y)=0 the joint equation of its asymptotes and the conjugate hyperbola respectively, then for any point (α,β)(\alpha, \beta) in the plane, H(α,β),A(α,β)H(\alpha, \beta), A(\alpha, \beta), and C(α,β)C(\alpha, \beta) are in
View answer and explanation
Correct option: A
A.P.
Option A (A.P.) is correct because the relationship between the hyperbola, its asymptotes, and the conjugate hyperbola results in an arithmetic progression.
285 Easy
285 of 298
Application +4 | -1
The line 42x−5y=404 \sqrt{2} x-5 y=40 touches the hyperbola x2100−y264=1\frac{x^2}{100}-\frac{y^2}{64}=1 at the point
View answer and explanation
Correct option: B
(102,8)(10 \sqrt{2}, 8)
Option B ((102,8)(10 \sqrt{2}, 8)) is correct because the point satisfies the equation of the tangent to the hyperbola at that point.
286 Easy
286 of 298
Application +4 | -1
The area of the triangle formed by the asymptotes and any tangent to the hyperbola x2−y2=a2x^2-y^2=a^2 is
View answer and explanation
Correct option: D
a2a^2
Option D is correct because the area of the triangle formed by the asymptotes and any tangent to the hyperbola is a2a^2.
287 Easy
287 of 298
Application +4 | -1
The line xcos⁡α+ysin⁡α=px \cos \alpha+y \sin \alpha=p touches the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, if
View answer and explanation
Correct option: A
a2cos⁡2α−b2sin⁡2α=p2a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2
Option A (a2cos⁡2α−b2sin⁡2α=p2a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2) is correct because it represents the condition for the line to be tangent to the hyperbola.
288 Easy
288 of 298
Application +4 | -1
If the chords of contact of the tangents from two points (x1,y1)\left(x_1, y_1\right) and (x2,y2)\left(x_2, y_2\right) to the hyperbola 4x2−5y2=a24 x^2-5 y^2=a^2 are at right angles. Then 16x1x2+25y1y216 x_1 x_2+25 y_1 y_2 is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because the condition for perpendicularity of the chords of contact leads to the equation 16x1x2+25y1y2=016 x_1 x_2+25 y_1 y_2=0.
289 Easy
289 of 298
Application +4 | -1
The equation of the hyperbola referred to its axes as axes of coordinates whose distance between the foci is 16 and eccentricity 2\sqrt{2}, is
View answer and explanation
Correct option: B
x2−y2=32x^2-y^2=32
Option B (x2−y2=32x^2-y^2=32) is correct because given the distance between foci and eccentricity, the equation of the hyperbola can be derived as x2−y2=32x^2-y^2=32.
290 Easy
290 of 298
Application +4 | -1
The equation to the hyperbola having its eccentricity 2 and the distance between its foci is 8, is
View answer and explanation
Correct option: B
x24−y212=1\frac{{{x^2}}}{4} - \frac{{{y^2}}}{{12}} = 1
Option B (x24−y212=1\frac{{{x^2}}}{4} - \frac{{{y^2}}}{{12}} = 1) is correct because it represents the equation of a hyperbola with the given eccentricity and distance between foci.
291 Easy
291 of 298
Application +4 | -1
The eccentricity of the hyperbola conjugate to x2−3y2=2x+8{x^2} - 3{y^2} = 2x + 8 is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the eccentricity of the conjugate hyperbola is calculated after finding the equation of the conjugate hyperbola.
292 Easy
292 of 298
Remembering +4 | -1
The curve xy=c2x y=c^2 is said to be
View answer and explanation
Correct option: B
A rectangular hyperbola
Option B (A rectangular hyperbola) is correct because the equation xy=c2xy = c^2 represents a rectangular hyperbola.
293 Easy
293 of 298
Application +4 | -1
The distance between the foci of the conic 7x2−9y2=637{x^2} - 9{y^2} = 63 is equal to
View answer and explanation
Correct option: A
8
Option A (8) is correct because it is the distance between the foci of the hyperbola, calculated using the formula 2ae.
294 Easy
294 of 298
Application +4 | -1
The diameter of 16x2−9y2=14416{x^2} - 9{y^2} = 144 which is conjugate to x=2yx = 2y is
View answer and explanation
Correct option: B
y=32x9y = \frac{{32x}}{9}
Option B (y=32x9y = \frac{{32x}}{9}) is correct because it represents the diameter conjugate to the given diameter of the hyperbola.
295 Easy
295 of 298
Understanding +4 | -1
The locus of the equation x2−y2=0,{x^2} - {y^2} = 0, is
View answer and explanation
Correct option: B
a hyperbola
Option C (a pair of lines) is correct because the equation x² - y² = 0 can be factored into (x - y)(x + y) = 0, representing two straight lines.
296 Easy
296 of 298
Understanding +4 | -1
The locus of a point which moves such that the difference of its distances from two fixed points is always a constant, is
View answer and explanation
Correct option: C
a hyperbola
Option D (Centre and vertices only) is correct because both the ellipse and hyperbola share the same center (0,0) and vertices (+-5, 0).
297 Easy
297 of 298
Analyzing +4 | -1
If tangent to any member of family of hyperbolas xy=4sin⁡2θ,θ∈(0,2π)−{π}x y=4 \sin ^2 \theta, \theta \in(0,2 \pi)-\{\pi \} is not a normal to any member of family of circles x2+y2−2x−2y+μ=0x^2+y^2-2 x-2 y+\mu=0, where μ\mu is any real parameter, then θ\theta belongs to
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the given condition holds true for all the specified intervals of θ\theta.
298 Easy
298 of 298
Application +4 | -1
The equation of the diameter which is conjugate to yy =3x=3 x w.r.t. the hyperbola x24−y29=1\frac{x^2}{4}-\frac{y^2}{9}=1, is
View answer and explanation
Correct option: A
y=34xy=\frac{3}{4} x
Option A is correct because the equation of the conjugate diameter is found using the relation m1m2=b2/a2m_1 m_2 = b^2/a^2.
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