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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 1-30 of 429 questions
1 Easy
1 of 429
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 Medium
2 of 429
Analyzing +4 | -1
Let P(x0,y0)P\left(x_{0}, y_{0}\right) be the point on the hyperbola 3x2−4y23 x^{2}-4 y^{2} =36=36, which is nearest to the line 3x+2y=13 x+2 y=1. Then 2(y0−x0)\sqrt{2}\left(y_{0}-x_{0}\right) is equal to
View answer and explanation
Correct option: C
−9-9
Option C (-9) is correct because it involves finding the point on the hyperbola where the tangent is parallel to the given line and then calculating the required expression.
3 Medium
3 of 429
Application +4 | -1
Suppose an ellipse and a hyperbola have the same pair of foci on the xx-axis with centres at the origin and they intersect at (2,2)(2,2). If the eccentricity of the ellipse is 12\frac{1}{2}, then the eccentricity of the hyperbola, is
View answer and explanation
Correct option: B
73\sqrt{\frac{7}{3}}
Option B (73\sqrt{\frac{7}{3}}) is correct because it is the eccentricity of the hyperbola derived from the given conditions of shared foci and intersection point.
4 Easy
4 of 429
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.
5 Easy
5 of 429
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.
6 Easy
6 of 429
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.
7 Easy
7 of 429
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.
8 Easy
8 of 429
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.
9 Easy
9 of 429
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.
10 Medium
10 of 429
Analyzing +4 | -1
Study the following statement and select the correct option:
A\mathrm{A} is the vertex of a branch of the hyperbola x2−2y2−22x−42y−6=0, Bx^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0, \mathrm{~B} is one of the end point of its latusrectum and C\mathrm{C} is the focus of the hyperbola nearer to the point AA
Statement-I : The area of the triangle ABC\mathrm{ABC} is 32−1\sqrt{\frac{3}{2}}-1

Statement-II: Eccentricity of the hyperbola is 222 \sqrt{2} and length of the conjugate axis is 4
View answer and explanation
Correct option: C
Statement-1 is true, statement-2 is false
Option C (Statement-1 is true, statement-2 is false) is correct because the area of the triangle is indeed 32−1\sqrt{\frac{3}{2}}-1, but the eccentricity and length of the conjugate axis are not as stated.
11 Medium
11 of 429
Application +4 | -1
The line lx+my+n=0l x+m y+n=0 is a tangent to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 if
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 the correct condition for tangency of a line to a hyperbola.
12 Easy
12 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. 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.
13 Medium
13 of 429
Application +4 | -1
Locus of intersection of the two perpendicular tangents to the given hyperbola is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the director circle equation results in a negative radius squared, indicating no real locus.
14 Easy
14 of 429
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.
15 Easy
15 of 429
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.

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16 Medium
16 of 429
Application +4 | -1
A tangent drawn to hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 at P(π6)P\left(\frac{\pi}{6}\right) forms a triangle of area 3a23 a^2 sq. units, with coordinate axes. Its eccentricity is equal to
View answer and explanation
Correct option: D
17\sqrt{17}
Option D is correct because the area of the triangle formed by the tangent is used to find the relationship between a and b, which is then used to calculate the eccentricity.
17 Easy
17 of 429
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.
18 Medium
18 of 429
Application +4 | -1
If e1{e_1} is the eccentricity of the ellipse x216+y225=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{25}} = 1 and e2{e_2} is the eccentricity of the hyperbola passing through the foci of the ellipse and e1e2=1{e_1}{e_2} = 1, then equation of the hyperbola is
View answer and explanation
Correct option: B
x216−y29=−1\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = - 1
Option B (x216−y29=−1\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = - 1) is correct because it satisfies the condition that the hyperbola passes through the foci of the ellipse and has the given eccentricity relationship.
19 Easy
19 of 429
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).
20 Easy
20 of 429
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.
21 Easy
21 of 429
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.
22 Easy
22 of 429
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.
23 Medium
23 of 429
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 x2+y2=9x^2+y^2=9, then minimum value of the expression 9x2+9y2\frac{9}{x^2}+\frac{9}{y^2} is equal to -
View answer and explanation
Correct option: A
4
The correct answer is A (4) because the minimum value of the expression 9x2+9y2\frac{9}{x^2} + \frac{9}{y^2} is found by substituting x=3cos⁡θx = 3\cos\theta and y=3sin⁡θy = 3\sin\theta and then using the fact that tan2θ+cot2θ>=2tan^2 \theta + cot^2 \theta >= 2.
24 Medium
24 of 429
Application +4 | -1
If distance between directrices of a rectangular hyperbola is 10, then distance between its foci will be
View answer and explanation
Correct option: D
20
Option D (20) is correct because, for a rectangular hyperbola, e = √2, and using the given distance between directrices, the distance between foci is calculated as 20.
25 Medium
25 of 429
Analyzing +4 | -1
Equation of asymptotes of xy=7x+5yxy = 7x + 5y are
View answer and explanation
Correct option: B
x=5,y=7x = 5,y = 7
Option B (x=5,y=7x = 5, y = 7) is correct because these are the asymptotes of the given hyperbola after rewriting the equation.
26 Easy
26 of 429
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.
27 Medium
27 of 429
Analyzing +4 | -1
The circle x2+y2−8x=0x^2+y^2-8 x=0 and hyperbola x29−y24=1\frac{x^2}{9}-\frac{y^2}{4}=1 intersect at the points AA and BB. The equation of a common tangent with positive slope to the circle as well as to the hyperbola, is
View answer and explanation
Correct option: B
2x−5y+4=02 x-\sqrt{5} y+4=0
Option B (2x−5y+4=02 x-\sqrt{5} y+4=0) is correct because it represents a common tangent with a positive slope to both the circle and the hyperbola.
28 Easy
28 of 429
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.
29 Easy
29 of 429
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.
30 Easy
30 of 429
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.
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