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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 361-390 of 429 questions

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361 Medium
361 of 429
Application +4 | -1
PP is a point on the hyperbola x281−y249=1\frac{x^2}{81}-\frac{y^2}{49}=1. The tangent at PP meets the transverse axis at T,NT, N is the foot of the perpendicular from PP to the transverse axis. If OO is the origin, then ON.OTON.OT is equal to.
View answer and explanation
Correct option: A
81
Option A (81) is correct because the product of ON and OT is calculated using the parametric coordinates of a point on the hyperbola and the equation of the tangent.
362 Easy
362 of 429
Application +4 | -1
Let x2a2+y2b2=1,a>b\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a > b be an ellipse, whose eccentricity is 12\frac{1}{\sqrt{2}} and the length of the latusrectum is 14\sqrt{14}. Then the square of the eccentricity of x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 is:
View answer and explanation
Correct option: A
32\frac{3}{2}
Option A (32\frac{3}{2}) is correct because it is the square of the eccentricity of the hyperbola calculated using the given information about the ellipse.
363 Medium
363 of 429
Application +4 | -1
Any point on the hyperbola (x+1)216−(y−2)24=1\frac{{{{\left( {x + 1} \right)}^2}}}{{16}} - \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1 is of the form
View answer and explanation
Correct option: D
(4  sec  θ−1,  2tan⁡θ+2)(4\mathrm{\;sec\;\theta } - 1,\mathrm{\;}2\tan \mathrm{\theta } + 2)
Option D ((4  sec;θ−1,  2tan⁡θ+2)(4\mathrm{\;sec\\;\theta } - 1,\mathrm{\;}2\tan \mathrm{\theta } + 2)) is correct because it represents a general point on the given hyperbola.
364 Easy
364 of 429
Application +4 | -1
The eccentricity of the conic represented by x2−y2−4x+4y+16=0x^2-y^2-4 x+4 y+16=0, is
View answer and explanation
Correct option: B
2\sqrt{2}
Option B (2\sqrt{2}) is correct because the given equation represents a rectangular hyperbola, which always has an eccentricity of 2\sqrt{2}.
365 Easy
365 of 429
Application +4 | -1
The locus of the poles of the chords of the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 which subtend a right angle at its centre is
View answer and explanation
Correct option: C
x2a4+y2b4=1a2−1b2\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2}
Option C (x2a4+y2b4=1a2−1b2\frac{{x^2}}{{a^4}}+\frac{{y^2}}{{b^4}}=\frac{{1}}{{a^2}}-\frac{{1}}{{b^2}}) is correct because the locus of the poles can be found by homogenizing the hyperbola equation using the polar equation and then applying the condition for perpendicular lines at the center.
366 Easy
366 of 429
Analyzing +4 | -1
Eccentricity of the hyperbola 2x2−2y2−3xy+9x+12y=02 x^2-2 y^2-3 x y+9 x+12 y=0 is equal to
View answer and explanation
Correct option: A
2\sqrt{2}
The correct answer is 2\sqrt{2} because the angle between the asymptotes is 90 degrees, indicating a rectangular hyperbola, which has an eccentricity of 2\sqrt{2}.
367 Easy
367 of 429
Application +4 | -1
The equation of the normal to the hyperbola x216−y29=1\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1 at (−4,0)\left( { - 4,0} \right) is
View answer and explanation
Correct option: D
y=0y = 0
Option D (y=0y = 0) is correct because the normal to a hyperbola at (-4,0) is a vertical line, and since the point lies on the x-axis, the normal is the x-axis itself.
368 Easy
368 of 429
Analyzing +4 | -1
For the hyperbola x2cos⁡2α−y2sin⁡2α=1\frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 1, which of the following remains constant when α\alpha varies?
View answer and explanation
Correct option: D
Abscissae of foci
Option D (Abscissae of foci) is correct because the x-coordinates of the foci remain constant at ±1\pm 1 as α\alpha varies.
369 Easy
369 of 429
Application +4 | -1
What is the slope of the tangent line drawn to the hyperbola xy=a(a≠0)x y=a(a \neq 0) at the point (a,1)(a, 1) ?
View answer and explanation
Correct option: B
−1/a-1 / a
Option B (-1/a) is correct because it is the derivative of the hyperbola equation evaluated at the specified point.
370 Medium
370 of 429
Application +4 | -1
The product of the perpendiculars from the foci on any tangent to the hyperbola x264−y29=1\frac{x^2}{64}-\frac{y^2}{9}=1 is
View answer and explanation
Correct option: B
9
Option B (9) is correct because the product of the perpendiculars from the foci to any tangent of the hyperbola is equal to b2b^2.
371 Medium
371 of 429
Analyzing +4 | -1
The locus of the point of intersection of the lines xcot⁡θ+y  cosecθ=2x\cot \mathrm{\theta } + y\;\mathrm{cosec \theta } = 2 and x  cosecθ+ycot⁡θ=6x\;\mathrm{cosec \theta } + y\cot \mathrm{\theta } = 6 is
View answer and explanation
Correct option: C
A hyperbola
Option C (A hyperbola) is correct because eliminating theta from the given equations results in the equation of a hyperbola.
372 Easy
372 of 429
Application +4 | -1
The locus of a point P(α,β)\mathrm{P}(\alpha, \beta) which is moving under the condition that the line y=αx+βy=\alpha x+\beta is a 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
A hyperbola
Option B (A hyperbola) is correct because the condition for tangency to a hyperbola results in an equation that represents another hyperbola.
373 Easy
373 of 429
Application +4 | -1
The locus of the point of intersection of the tangents to the hyperbola x216−y236=1\frac{x^2}{16}-\frac{y^2}{36}=1 which are at right angles is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the locus is a circle x2+y2=16−36x^2+y^2=16-36 which is not possible, so there is no pair of tangents that are at right angles.
374 Easy
374 of 429
Application +4 | -1
The centre of the hyperbola 9x2−36x−16y2+96y−252=09 x^2-36 x-16 y^2+96 y-252=0, is
View answer and explanation
Correct option: A
(2,3)(2,3)
Option A ((2,3)(2,3)) is correct because the center is directly identified from the standard form of the hyperbola equation after completing the square.
375 Easy
375 of 429
Application +4 | -1
Acceleration of a particle moving along a straight line is a function of velocity as a=2va=2 \sqrt{v}. At t=2st=2 s, its velocity v=16 ms−1v=16 \mathrm{~ms}^{-1}. Its velocity at t=3t=3 s will be
View answer and explanation
Correct option: B
25 ms−125 \mathrm{~ms}^{-1}
Option B (25 ms−125 \mathrm{~ms}^{-1}) is correct because solving the differential equation a=2va = 2\sqrt{v} with the given initial condition yields v=(t+2)2v = (t+2)^2, and at t=3, v=25.
376 Medium
376 of 429
Application +4 | -1
The normal at P(x1,y1)P\left(x_1, y_1\right) on the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 meets the coordinate axes at AA and BB. If OO, is the origin and ee, the eccentricity of the hyperbola, then
View answer and explanation
Correct option: A
OA=e2x1O A=e^2 x_1
Option A (OA=e2x1OA=e^2 x_1) is correct because the x-intercept of the normal is e2x1e^2x_1, which represents the distance OA.
377 Medium
377 of 429
Application +4 | -1
Any point on the hyperbola (x+1)216−(y−2)24=1\frac{{{{\left( {x + 1} \right)}^2}}}{{16}} - \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1 is of the form
View answer and explanation
Correct option: B
(4sec⁡θ−1,2tan⁡θ+2)(4\sec \mathrm{\theta } - 1,2\tan \mathrm{\theta } + 2)
Option B ((4sec⁡θ−1,2tan⁡θ+2)(4\sec \mathrm{\theta } - 1,2\tan \mathrm{\theta } + 2)) is correct because it represents a general point on the given hyperbola after considering the shift in the center.
378 Easy
378 of 429
Application +4 | -1
The co-ordinate of focus of 9x2−16y2+18x+32y−151=09 x^2-16 y^2+18 x+32 y-151=0
View answer and explanation
Correct option: A
(−6,1)(-6,1)
Option A ((-6,1)) is correct because completing the square and transforming the equation into standard form reveals the center and allows calculation of the foci coordinates.
379 Easy
379 of 429
Application +4 | -1
If the line lx+my+n=0l x+m y+n=0 touches the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, then
View answer and explanation
Correct option: B
a2l2−b2m2=n2a^2 l^2-b^2 m^2=n^2
Option B (a2l2−b2m2=n2a^2 l^2-b^2 m^2=n^2) is correct because it represents the condition for the line to be tangent to the hyperbola.
380 Easy
380 of 429
Analyzing +4 | -1
The locus of the point of intersection of the tangents at the end-points of normal chords of the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, is
View answer and explanation
Correct option: B
a6x2−b6y2=(a2+b2)2\frac{a^6}{x^2}-\frac{b^6}{y^2}=\left(a^2+b^2\right)^2
Option B (a6x2−b6y2=(a2+b2)2\frac{a^6}{x^2}-\frac{b^6}{y^2}=\left(a^2+b^2\right)^2) is correct because it represents the locus of the point of intersection of tangents at the end-points of normal chords of the hyperbola.
381 Medium
381 of 429
Application +4 | -1
The condition that the straight line lx+my=nl x+m y=n may be a normal to the hyperbola b2x2−a2y2=a2b2b^2 x^2-a^2 y^2=a^2 b^2 is given by
View answer and explanation
Correct option: A
a2l2−b2m2=(a2+b2)2n2\frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{\left(a^2+b^2\right)^2}{n^2}
Option A (a2l2−b2m2=(a2+b2)2n2\frac{{a^2}}{{l^2}}-\frac{{b^2}}{{m^2}}=\frac{{(a^2+b^2)^2}}{{n^2}}) is correct because it is the condition derived by comparing the general normal equation to the hyperbola with the given line equation.
382 Easy
382 of 429
Analyzing +4 | -1
Let the foci of the ellipse x216+y27=1\frac{x^2}{16}+\frac{y^2}{7}=1 and the hyperbola x2144−y2α=125\frac{x^2}{144}-\frac{y^2}{\alpha}=\frac{1}{25} coincide. Then the length of the latus rectum of the hyperbola is:
View answer and explanation
Correct option: D
2710\frac{27}{10}
Option D (2710\frac{27}{10}) is correct because the length of the latus rectum is calculated using the formula 2b2/a2b^2/a after finding the value of α\alpha by equating the foci of the ellipse and hyperbola.
383 Easy
383 of 429
Application +4 | -1
The coordinates of the centre of the hyperbola x2+3xy+2y2+2x+3y+2=0{x^2} + 3xy + 2{y^2} + 2x + 3y + 2 = 0
View answer and explanation
Correct option: A
(−1,0)(-1,0)
Option A ((-1,0)) is correct because the center of the hyperbola is found by solving the partial derivatives of the equation with respect to x and y, setting them equal to zero.
384 Easy
384 of 429
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.
385 Easy
385 of 429
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.
386 Easy
386 of 429
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.
387 Easy
387 of 429
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.
388 Medium
388 of 429
Application +4 | -1
Length of the straight line x−3y=1x - 3y = 1 intercepted by the hyperbola x2−4y2=1{x^2} - 4{y^2} = 1 is
View answer and explanation
Correct option: B
6510\frac{6}{5}\sqrt {10}
Option B (6510\frac{6}{5}\sqrt {10}) is correct because it is the distance between the intersection points of the line and the hyperbola.
389 Medium
389 of 429
Application +4 | -1
The value of mm, for which the line y=mx+2533y=m x+\frac{25 \sqrt{3}}{3} is a normal to the conic x216−y29=1\frac{x^2}{16}-\frac{y^2}{9}=1, is
View answer and explanation
Correct option: B
−2/3-2 / \sqrt{3}
Option D (2x+3y=252 x+\sqrt{3} y=25) is correct because substituting the point (8, 3√3) into the equation satisfies it, and it is the normal to the hyperbola at that point.
390 Easy
390 of 429
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.
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