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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 1-30 of 298 questions
1 Easy
1 of 298
Application +4 | -1
Locus of the point of intersection of the tangents at the points with eccentric angles ϕ\phi and π2−ϕ\frac{\pi}{2}-\phi on the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 is
View answer and explanation
Correct option: B
y=by=b
Option B (y=by=b) is correct because the locus of the point of intersection of tangents at the given eccentric angles simplifies to y=b after substituting the angles into the intersection point formula.
2 Easy
2 of 298
Application +4 | -1
The eccentricity of the hyperbola 9x2−16y2+72x−32y−16=09 x^2-16 y^2+72 x-32 y-16=0, is
View answer and explanation
Correct option: A
54\frac{5}{4}
Option A (54\frac{5}{4}) is correct because it correctly completes the square to find the standard form of the hyperbola and then calculates the eccentricity.
3 Easy
3 of 298
Analyzing +4 | -1
Let the hyperbola H:x2a2−y2b2=1H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 pass through the point (22,−22)(2 \sqrt{2},-2 \sqrt{2}). A parabola is drawn whose focus is same as the focus of HH with positive abscissa and the directrix of the parabola passes through the other focus of HH. If the length of the latus rectum of the parabola is ee times the length of the latus rectum of HH, where ee is the eccentricity of H\mathrm{H}, then which of the following points lies on the parabola?
View answer and explanation
Correct option: B
(33,−62)(3 \sqrt{3},-6 \sqrt{2})
Option B ((33,−62)(3 \sqrt{3},-6 \sqrt{2})) is correct because it satisfies the equation of the parabola y2=83xy^2 = 8\sqrt{3}x after determining the values of a, b, and e.
4 Easy
4 of 298
Application +4 | -1
The equation of the conic with focus at (1,−1),\left( {1, - 1} \right), directrix along x−y+1=0x - y + 1 = 0 and with eccentricity 2\sqrt 2 is
View answer and explanation
Correct option: C
2xy−4x+4y+1=02xy - 4x + 4y + 1 = 0
Option C (2xy−4x+4y+1=02xy - 4x + 4y + 1 = 0) is the correct conic equation because it satisfies the definition of a conic section with the given focus, directrix, and eccentricity.
5 Easy
5 of 298
Application +4 | -1
Distance between foci is 8 and distance between directrices is 6 of hyperbola, then length of latusrectum is
View answer and explanation
Correct option: B
43\frac{4}{{\sqrt 3 }}
Option B (43\frac{4}{{\sqrt 3 }}) is correct because it is the result of calculating the latus rectum using the given information about the distance between foci and directrices to find a and b.
6 Easy
6 of 298
Application +4 | -1
The equation of the hyperbola, whose foci are the foci of the ellipse x225+y29=1\frac{x^2}{25}+\frac{y^2}{9}=1 and the eccentricity is 2 , 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 the foci of the ellipse are found, then used to determine the equation of the hyperbola with the given eccentricity.
7 Easy
7 of 298
Application +4 | -1
A hyperbola passes through the foci of the ellipse x225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1 and the product of their eccentricities is 1. The equation of the hyperbola is
View answer and explanation
Correct option: B
x29−y216=1\frac{x^2}{9}-\frac{y^2}{16}=1
Option B (x29−y216=1\frac{{x^2}}{{9}}-\frac{{y^2}}{{16}}=1) is correct because it satisfies the given conditions: passing through the foci of the ellipse and having the product of eccentricities equal to 1.
8 Easy
8 of 298
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. If the eccentricity of the hyperbola is 2 , then the equation of the tangent of this hyperbola passing through the point (4,6)(4,6) is
View answer and explanation
Correct option: A
2x−y−2=02 x-y-2=0
Option A (2x−y−2=02x - y - 2 = 0) is correct because by finding the foci of the ellipse, then using the eccentricity of the hyperbola, we can determine the hyperbola's equation and subsequently find the tangent at the given point.
9 Easy
9 of 298
Application +4 | -1
The hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 has its conjugate axis 5 and passes through the point (2,−1)(2,-1). The length of a latus rectum is
View answer and explanation
Correct option: D
5429\frac{5}{4} \sqrt{29}
Option D (5429\frac{5}{4} \sqrt{29}) is correct because it is the length of the latus rectum calculated using the given information about the hyperbola.
10 Easy
10 of 298
Application +4 | -1
The equation of the chord of x2−y2=9x^2-y^2=9 which is bisected at (5,−3)(5,-3), is
View answer and explanation
Correct option: A
5x+3y−16=05 x+3 y-16=0
Option A (5x+3y−16=05 x+3 y-16=0) is correct because it represents the equation of the chord bisected at the given point, found using the T = S' method.
11 Easy
11 of 298
Application +4 | -1
A Basic Terms of Conics is defined by the equations x=−1+sec⁡t,  y=2+3tan⁡t.x = - 1 + \sec t,\;y = 2 + 3\tan t. The coordinates of the foci are
View answer and explanation
Correct option: A
(−1−10,2)\left( { - 1 - \sqrt {10} ,2} \right) and (−1+10,2)\left( { - 1 + \sqrt {10} ,2} \right)
Option A is correct because the foci of the hyperbola are calculated using the eccentricity and the center of the hyperbola derived from the given parametric equations.
12 Easy
12 of 298
Remembering +4 | -1
The length of latusrectum of hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 equals
View answer and explanation
Correct option: B
2a(e2−1)2a\left( {{e^2} - 1} \right)
Option B (2a(e2−1)2a\left( {{e^2} - 1} \right)) is correct because the length of the latus rectum of a hyperbola is given by 2b2a\frac{2b^2}{a}, which can be expressed as 2a(e2−1)2a(e^2 - 1).
13 Easy
13 of 298
Application +4 | -1
The eccentricity of the hyperbola x216−y225=1\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{{25}} = 1 is
View answer and explanation
Correct option: C
414\frac{{\sqrt {41} }}{4}
Option C (414\frac{{\sqrt {41} }}{4}) is correct because it is the eccentricity calculated using the formula for a hyperbola.
14 Easy
14 of 298
Application +4 | -1
The difference between the length 2a2 a of the transverse axis of a hypertola of eccentricity ee and the length of its latus rectum is
View answer and explanation
Correct option: C
2a∣e2−1∣2 a | e^2-1|
Option C (2a∣e2−1∣2 a | e^2-1|) is correct because it represents the difference between the transverse axis and the latus rectum of the hyperbola.
15 Easy
15 of 298
Application +4 | -1
If x2+y2=k2x^2+y^2=k^2, (where xx and yy are variables and kk is constant), then x&yx \& y can be taken as x=kcos⁡θx=k \cos \theta &y=ksin⁡θ\& y=k \sin \theta. If (x−1)2+(y−2)2=4(x-1)^2+(y-2)^2=4, then maximum value of x2+y2x^2+y^2 is equal to
View answer and explanation
Correct option: C
9+459+4 \sqrt{5}
The correct answer is C (9+459+4 \sqrt{5}) because it represents the maximum value of x2+y2x^2 + y^2 when x and y are parameterized using trigonometric functions and the maximum is found using the range of acos⁡θ+bsin⁡θa \cos \theta + b \sin \theta.

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16 Easy
16 of 298
Application +4 | -1
If the eccentricity of the standard hyperbola passing through the point (4, 6) is 2, then the equation of the tangent to the hyperbola at (4, 6) is
View answer and explanation
Correct option: C
2x−y−2=02x - y - 2 = 0
Option C (2x−y−2=02x - y - 2 = 0) is correct because it represents the tangent to the hyperbola at the point (4, 6) given the eccentricity is 2.
17 Easy
17 of 298
Application +4 | -1
The line 3x+2y+1=03 x+2 y+1=0 meets the hyperbola 4x2−y24 x^2-y^2 =4a2=4 a^2 in the points P{P} and Q{Q}. The coordinates of the point of intersection of the tangents at P{P} and Q{Q} are
View answer and explanation
Correct option: A
(−3a2,8a2)\left(-3 a^2, 8 a^2\right)
Option A is correct because the point of intersection of tangents is the pole of the given line with respect to the hyperbola.
18 Easy
18 of 298
Application +4 | -1
Length of the latus rectum of the hyperbola x2tan⁡α−y2cot⁡α=1,(0<α<π/4)x^2 \tan \alpha-y^2 \cot \alpha=1,(0 < \alpha < \pi / 4) is
View answer and explanation
Correct option: C
2(tan⁡α)3/22(\tan \alpha)^{3 / 2}
Option C (2(tan⁡α)3/22(\tan \alpha)^{3 / 2}) is correct because it is the correct formula for the length of the latus rectum of the given hyperbola.
19 Easy
19 of 298
Application +4 | -1
Asymptotes of a hyperbola x225−y216=1\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{16}} = 1 are
View answer and explanation
Correct option: D
y=±45xy = \pm \frac{4}{5}x
Option D (y=±45xy = \pm \frac{4}{5}x) is correct because it represents the asymptotes of the given hyperbola, derived from the standard form equation.
20 Easy
20 of 298
Application +4 | -1
The product of the lengths of the perpendiculars drawn from 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: A
b2b^2
Option A (b2b^2) is correct because the product of the lengths of the perpendiculars from the foci to any tangent of the hyperbola is equal to b2b^2.
21 Easy
21 of 298
Application +4 | -1
If the latusrectum of the hyperbola x216−y2b2=1\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{{{b^2}}} = 1 is 92\frac{9}{2}, then its eccentricity is
View answer and explanation
Correct option: B
54\frac{5}{4}
Option B (54\frac{5}{4}) is correct because it is the eccentricity calculated from the given latus rectum of the hyperbola.
22 Easy
22 of 298
Application +4 | -1
Consider the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. Area of the triangle formed by the asymptotes and the tangent drawn to it at (a,0)(a, 0) is
View answer and explanation
Correct option: B
ab
Option B (ab) is correct because it represents the area of the triangle formed by the asymptotes and the tangent at the vertex of the hyperbola.
23 Easy
23 of 298
Analyzing +4 | -1
The curve represented by x=a(cosh⁡θ+sinh⁡θ),y=b(cosh⁡θ−sinh⁡θ),x = a\left( {\cosh \theta + \sinh \theta } \right),y = b\left( {\cosh \theta - \sinh \theta } \right), is
View answer and explanation
Correct option: A
A hyperbola
Option A (A hyperbola) is correct because the equation simplifies to xy = ab, which represents a hyperbola.
24 Easy
24 of 298
Analyzing +4 | -1
For each positive integer nn, consider the point PP with abscissa nn on the curve y2−x2=1y^2-x^2=1. If dnd_n represents the shortest distance from the point PP to the line y=xy=x, then lim⁡n→∞(n.dn)\lim\limits_{n \to \infty } \left( {n.{d_n}} \right) has the value equal to
View answer and explanation
Correct option: A
122\frac{1}{2 \sqrt{2}}
Option A (122\frac{1}{2 \sqrt{2}}) is correct because it is the limit of n⋅dnn \cdot d_n as nn approaches infinity, calculated using algebraic manipulation and limit properties.
25 Easy
25 of 298
Analyzing +4 | -1
Let a line L:2x+y=k,  k>0L:2x + y = k,\,\,k > 0 be a tangent to the hyperbola x2−y2=3{x^2} - {y^2} = 3. If LL is also a tangent to the parabola y2=αx{y^2} = \alpha x, then α\alpha is equal to
View answer and explanation
Correct option: D
-24
Option D (-24) is correct. The condition for tangency to the hyperbola gives the value of k, and then the condition for tangency to the parabola gives the value of α.
26 Easy
26 of 298
Application +4 | -1
The equation of the line passing through the centre of a rectangular hyperbola is x−y−1=0x-y-1=0.
If one of its asymptote is 3x−4y−6=03 x-4 y-6=0, the equation of the other asymptote is
View answer and explanation
Correct option: B
4x+3y+17=04 x+3 y+17=0
Option B (4x+3y+17=04 x+3 y+17=0) is correct because the asymptotes of a rectangular hyperbola are perpendicular, and the center lies on the given line. By finding the center and using the perpendicularity condition, we can determine the equation of the other asymptote.
27 Easy
27 of 298
Application +4 | -1
If mm is a variable, the locus of the point of intersection of the lines x3−y2=m\frac{x}{3}-\frac{y}{2}=m and x3+y2=1m\frac{x}{3}+\frac{y}{2}=\frac{1}{m}, is a/an
View answer and explanation
Correct option: C
Hyperbola
Option C (Hyperbola) is correct because the equation derived by eliminating the parameter mm results in the standard form of a hyperbola.
28 Easy
28 of 298
Analyzing +4 | -1
If the circle x2+y2=a2{x^2} + {y^2} = {a^2} intersects the hyperbola xy=c2xy = {c^2} in four points (xi,yi),\left( {{x_i},{y_i}} \right), for i=1,2,3i = 1,2,3 and 4, then y1+y2+y3+y4{y_1} + {y_2} + {y_3} + {y_4} equals
View answer and explanation
Correct option: A
0
Option A (0) is correct because the sum of the y-coordinates of the intersection points is zero, as derived from the equation of the circle and hyperbola.
29 Easy
29 of 298
Application +4 | -1
If the normal to the rectangular hyperbola xy=c2x y=c^2 at the point (ct,c/t)(c t, c / t) meets the curve again at (ct′,c/t′)\left(c t^{\prime}, c / t^{\prime}\right), then
View answer and explanation
Correct option: B
t3t′=−1t^3 t^{\prime}=-1
Option B (t3t′=−1t^3 t^{\prime}=-1) is correct because it is derived from the equation of the normal to the hyperbola and the condition that it intersects the curve again at a different point.
30 Easy
30 of 298
Application +4 | -1
The angle between pair of tangents to the curve 7x2−12y2=847 x^2-12 y^2=84 from the point M(1,2)M(1,2), is
View answer and explanation
Correct option: C
2(tan⁡−113+tan⁡−112)2\left(\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{2}\right)
Option C 2(tan⁡−113+tan⁡−112)2\left(\tan ^{{-1}} \frac{{1}}{{3}}+\tan ^{{-1}} \frac{{1}}{{2}}\right) is correct because the point lies on the director circle, implying a right angle between tangents, and the expression simplifies to π2\frac{{\pi}}{{2}}.
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