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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 421-429 of 429 questions

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421 Medium
421 of 429
Application +4 | -1
If a hyperbola passing through the origin has 3x+4y−1=03 x+4 y-1=0 and 4x−3y−6=04 x-3 y-6=0 as its assumptions, then the equation of the transverse axis of the hyperbola is
View answer and explanation
Correct option: B
x−7y−5=0x-7 y-5=0
Option B (x−7y−5=0x-7y-5=0) is correct because the transverse axis is the bisector of the angle between the asymptotes that contains the origin.
422 Easy
422 of 429
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.
423 Hard
423 of 429
Understanding +4 | -1
If a diameter of a hyperbola meets the hyperbola in real points then
View answer and explanation
Correct option: A
It meets the conjugate hyperbola in imaginary points.
The correct answer is A (It meets the conjugate hyperbola in imaginary points.) because if a diameter intersects a hyperbola in real points, it intersects its conjugate in imaginary points.
424 Medium
424 of 429
Analyzing +4 | -1
If a variable straight line xcos⁡α+ysin⁡α=px \cos \alpha+y \sin \alpha=p which is a chord of hyperbola x2a2−y2b2=1(b>a)\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(b>a) subtends a right angle at the centre of the hyperbola, then it always touches a fixed circle whose radius, is
View answer and explanation
Correct option: C
abb2−a2\frac{a b}{\sqrt{b^2-a^2}}
Option C (abb2−a2\frac{a b}{\sqrt{b^2-a^2}}) is correct because it represents the radius of the circle that the variable straight line always touches when it subtends a right angle at the center of the hyperbola.
425 Easy
425 of 429
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).
426 Medium
426 of 429
Application +4 | -1
The equation of the chord of the hyperbola 25x2−16y2=40025 x^2-16 y^2=400 that is bisected at point (5,3)(5,3) is:
View answer and explanation
Correct option: B
125x−48y=481125 x-48 y=481
Option B (125x−48y=481125 x-48 y=481) is correct because it is the equation of the chord of the hyperbola bisected at the point (5,3), found using the formula T=S1.
427 Medium
427 of 429
Application +4 | -1
If the tangent at the point P(asec⁡θ,btan⁡θ)P(a \sec \theta, b \tan \theta) to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 passes through the point where a directrix of the hyperbola meets the positive side of the transverse axis, then θ\theta is equal to
Where ee is the eccentricity of the hyperbola.
View answer and explanation
Correct option: A
cos⁡−1(1/e)\cos ^{-1}(1 / e)
Option A (cos⁡−1(1/e)\cos^{-1}(1/e)) is correct because substituting the directrix point into the tangent equation and solving for θ\theta gives cos⁡θ=1/e\cos \theta = 1/e.
428 Easy
428 of 429
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.
429 Easy
429 of 429
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.
Revision notes and key takeaways

Review HYPERBOLA in your textbook or class notes, then use this set to check your understanding.

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