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JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Medium

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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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31 Medium
31 of 118
Application +4 | -1
The circle described on the line joining the foci of the hyperbola x216−y29=1\frac{x^2}{16}-\frac{y^2}{9}=1 as a diameter passes through an end of the latus rectum of the parabola y2=4axy^2=4 a x, the length of the latus rectum of the parabola is
View answer and explanation
Correct option: C
454 \sqrt{5} units
Option C (454 \sqrt{5} units) is correct because finding the foci, circle equation, and parabola's latus rectum end, then solving for 'a' and substituting into 4a gives 454\sqrt{5}.
32 Medium
32 of 118
Application +4 | -1
If the chords of contact of tangents from two points (x1,  y1)  and  (x2,  y2)\left( {{x_1},\;{y_1}} \right)\mathrm{\;and}\;\left( {{x_2},\;{y_2}} \right) to the hyperbola 4x2−9y2−36=04{x^2} - 9{y^2} - 36 = 0 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
−8116- \frac{{81}}{{16}}
The correct answer is D (−8116- \frac{{81}}{{16}}) because it represents the ratio of the product of the x-coordinates to the product of the y-coordinates of the points, given that their chords of contact are at right angles.
33 Medium
33 of 118
Analyzing +4 | -1
A hyperbola touches yy-axis and has its centre and one of the foci are (52,20)\left(\frac{5}{2}, 20\right) and (10,24)(10,24) respectively, then length of its transverse axis is.
View answer and explanation
Correct option: C
89\sqrt{89}
The correct answer is C. By using the given center and focus, and the fact that the hyperbola touches the y-axis, the length of the transverse axis is calculated to be 89\sqrt{89}.
34 Medium
34 of 118
Application +4 | -1
Let aa and bb respectively be the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity ee satisfies the equation 9e2−18e+5=09 e^2-18 e+5=0. If S(5,0)S(5,0) is a focus and 5x=95 x=9 is the corresponding directrix of hyperbola, then a2−b2a^2-b^2 is equal to
View answer and explanation
Correct option: A
−7-7
Option A (-7) is correct because it correctly uses the given information to find a and b, then calculates a2 - b2.
35 Medium
35 of 118
Application +4 | -1
If α,β\alpha, \beta are the values of kk for which the foci of ellipse x2k2a2+y2a2=1\frac{x^2}{k^2 a^2}+\frac{y^2}{a^2}=1 and hyperbola x2a2−y2a2=1\frac{x^2}{a^2}-\frac{y^2}{a^2}=1 coincide, then the value of (α+β)2(\alpha+\beta)^2 is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the foci of the ellipse and hyperbola coincide only when k = ±√3, making α + β = 0, and thus (α + β)² = 0.
36 Medium
36 of 118
Application +4 | -1
The value of mm, for which the line y=mx+2533y = mx + \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: A
±23\pm \frac{2}{{\sqrt 3 }}
Option A (±23\pm \frac{2}{{\sqrt 3 }}) is correct because it satisfies the condition for the line to be a normal to the given hyperbola.
37 Medium
37 of 118
Application +4 | -1
If the tangent and normal to a rectangular hyperbola cut off intercepts x1x_1 and x2x_2 on one axis and y1y_1 and y2y_2 on the other axis, then
View answer and explanation
Correct option: C
x1x2+y1y2=0x_1 x_2+y_1 y_2=0
Option A (a1a2+b1b2=0a_1 a_2+b_1 b_2=0) is correct because the product of the intercepts on one axis plus the product of the intercepts on the other axis equals zero for a rectangular hyperbola.
38 Medium
38 of 118
Application +4 | -1
If eccentricities of hyperbolas x2a2−y2 b2=1\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 and y2 b2−x2a2=1\frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1 are e\mathrm{e} and e1\mathrm{e} 1 then 1e12+1e2=\frac{1}{\mathrm{e}_1^2}+\frac{1}{\mathrm{e}^2}=
View answer and explanation
Correct option: A
1
Option A (1) is correct because the eccentricities of the two hyperbolas are related such that 1e12+1e2=1\frac{1}{e_1^2} + \frac{1}{e^2} = 1.
39 Medium
39 of 118
Application +4 | -1
If the tangent at any point PP on the hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 meets the lines bx−ay=0bx - ay = 0 and bx+ay=0bx + ay = 0 in the points QQ and RR, then CQ.CR=CQ.CR =
View answer and explanation
Correct option: C
a2+b2{a^2} + {b^2}
Option C (a2+b2{a^2} + {b^2}) is correct because it represents the product of the distances from the origin to the points where the tangent intersects the given lines.
40 Medium
40 of 118
Application +4 | -1
P{P} is a variable point on the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 whose vertex is A(a,0){A}(a, 0). The locus of the middle point of APAP is
View answer and explanation
Correct option: B
(2x−a)2a2−4y2b2=1\frac{(2 x-a)^2}{a^2}-\frac{4 y^2}{b^2}=1
Option B is correct because the locus of the midpoint is found by using the equation of the chord with a given midpoint.
41 Medium
41 of 118
Application +4 | -1
The length of the latusrectum of the hyperbola xy−3x−3y+7=0xy - 3x - 3y + 7 = 0 is
View answer and explanation
Correct option: B
4
Option B (4) is correct because the length of the latus rectum of the given rectangular hyperbola is 4.
42 Medium
42 of 118
Application +4 | -1
Let H:−x2a2+y2 b2=1\mathrm{H}: \frac{-\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 be the hyperbola, whose eccentricity is 3\sqrt{3} and the length of the latus rectum is 434 \sqrt{3}. Suppose the point (α,6),α>0(\alpha, 6), \alpha > 0 lies on HH. If β\beta is the product of the focal distances of the point (α,6)(\alpha, 6), then α2+β\alpha^2+\beta is equal to :
View answer and explanation
Correct option: B
171
Option B (171) is correct because by finding the equation of the hyperbola and the focal distances, we can calculate α2+β\alpha^2 + \beta to be 171.
43 Medium
43 of 118
Analyzing +4 | -1
If chords of the hyperbola x2−y2=a2{x^2} - {y^2} = {a^2} touch the parabola y2=4ax.{y^2} = 4ax. Then, the locus of the middle points of these chord is
View answer and explanation
Correct option: B
y2(x−a)=x3{y^2}\left( {x - a} \right) = {x^3}
Option B (y2(x−a)=x3{y^2}({x - a}) = {x^3}) is correct because it represents the locus of the midpoints of the chords of the hyperbola that touch the given parabola.
44 Medium
44 of 118
Application +4 | -1
What is the equation of the chord of hyperbola 25x2−16y2=40025 x^2-16 y^2=400 whose mid-point is (5,3)(5,3) ?
View answer and explanation
Correct option: B
125x−48y=481125 x-48 y=481
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 for the line to be a normal to the hyperbola, derived by comparing coefficients.
45 Medium
45 of 118
Analyzing +4 | -1
Equation 1r=18+38cos⁡θ\frac{1}{r} = \frac{1}{8} + \frac{3}{8}\cos \mathrm{\theta } represents
View answer and explanation
Correct option: B
A hyperbola
Option B (A hyperbola) is correct because the given polar equation has an eccentricity greater than 1, indicating a hyperbola.
46 Medium
46 of 118
Application +4 | -1
The line lx+my+n=0l x+m y+n=0 will be a normal to the hyperbola b2x2−a2y2=a2b2b^2 x^2-a^2 y^2=a^2 b^2, if
View answer and explanation
Correct option: C
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 C (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}) is correct because it represents the condition for the line to be a normal to the hyperbola.
47 Medium
47 of 118
Application +4 | -1
Let the eccentricity of the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 be the reciprocal to that of the ellipse x2+4y2=4x^2+4 y^2=4. If the hyperbola passes through a focus of the ellipse, then the equation of the hyperbola, is
View answer and explanation
Correct option: B
x2−3y2=3x^2-3 y^2=3
Option B (x2−3y2=3x^2-3 y^2=3) is correct because the hyperbola passes through a focus of the ellipse and has the reciprocal eccentricity.
48 Medium
48 of 118
Application +4 | -1
The coordinates of a point common to a directrix and an asymptote of the hyperbola x225−y216=1\frac{x^2}{25}-\frac{y^2}{16}=1 are
View answer and explanation
Correct option: D
(−2541,2041)\left(\frac{-25}{\sqrt{41}}, \frac{20}{\sqrt{41}}\right)
Option D ((−2541,2041)\left(\frac{{-25}}{{\sqrt{41}}}, \frac{{20}}{{\sqrt{41}}}\right)) is correct because it is the intersection point of the directrix and asymptote of the given hyperbola.
49 Medium
49 of 118
Application +4 | -1
A tangent to a hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 intercepts a length of unity from each of the coordinate axes, then the point (a,b)(a, b) lies on the rectangular hyperbola
View answer and explanation
Correct option: B
x2−y2=1x^2-y^2=1
Option B (x2−y2=1x^2-y^2=1) is correct because the tangent intercepts unity from each axis, leading to the condition that a=sec⁡θa = \sec \theta and b=tan⁡θb = \tan \theta, which satisfies the equation x2−y2=1x^2 - y^2 = 1.
50 Medium
50 of 118
Application +4 | -1
If the tangent on the point (2sec⁡ϕ,3tan⁡ϕ)(2 \sec \phi, 3 \tan \phi) of the hyperbola x24−y29=1\frac{x^2}{4}-\frac{y^2}{9}=1 is parallel to 3x−y+4=03 x-y+4=0, then the value of ϕ\phi is
View answer and explanation
Correct option: C
30∘30^{\circ}
Option C (30∘30^{\circ}) is correct because it is the angle that satisfies the condition where the tangent to the hyperbola is parallel to the given line.
51 Medium
51 of 118
Application +4 | -1
The equation of the hyperbola, referred to its axes as axes of coordinates, given that the distances of one of its vertices from the foci are 9 and 1 units, is
View answer and explanation
Correct option: A
x216−y29=1\frac{x^2}{16}-\frac{y^2}{9}=1
Option A is correct because the equation of the hyperbola is found using the given distances of the vertex from the foci.
52 Medium
52 of 118
Application +4 | -1
Tangent drawn from the point (c,d)(c, d) to the hyperbola 9x2−64y2=5769 x^2-64 y^2=576 make angle α\alpha and β\beta with the xx-axis. If tan⁡αtan⁡β=1\tan \alpha \tan \beta=1, then the value of c2−d2c^2-d^2 is equal to
View answer and explanation
Correct option: A
73
Option A (73) is correct because by substituting the point (c, d) into the tangent equation and using the given condition tan⁡αtan⁡β=1\tan \alpha \tan \beta = 1, we can solve for c2−d2c^2 - d^2.
53 Medium
53 of 118
Application +4 | -1
If a rectangular hyperbola (x−1)(y−2)=4(x-1)(y-2)=4 cuts a circle x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0 at points (3,4)(5,3)(3,4)(5,3), (2,6)(2,6) and (−1,0)(-1,0), then the value of (g+f)(g+f) is equal to
View answer and explanation
Correct option: A
-8
Option A (-8) is correct because the arithmetic mean of the intersection points of the circle and rectangular hyperbola is the midpoint of the centers of the hyperbola and circle, allowing us to solve for g+f.
54 Medium
54 of 118
Application +4 | -1
The length of the semi-transverse axis of the rectangular hyperbola xy=32xy = 32 is
View answer and explanation
Correct option: D
8
Option D (8) is correct because the semi-transverse axis of a rectangular hyperbola xy=c2xy=c^2 is given by a=2ca = \sqrt{2}c, and in this case, c2=32c^2 = 32.
55 Medium
55 of 118
Application +4 | -1
The asymptote of the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 form with any tangent to the hyperbola a triangle whose area is a2tan⁡λa^2 \tan \lambda in magnitude, then its eccentricity is
View answer and explanation
Correct option: A
sec⁡λ\sec \lambda
Option A (sec⁡λ\sec \lambda) is correct because it represents the eccentricity of the hyperbola, derived from the relationship between the area of the triangle formed by the asymptotes and a tangent, and the hyperbola's parameters.
56 Medium
56 of 118
Application +4 | -1
If e1{e_1} and e2{e_2} are the eccentricities of a hyperbola 3x2−3y2=253{x^2} - 3{y^2} = 25 and its conjugate, then
View answer and explanation
Correct option: B
e12+e22=4e_1^2 + e_2^2 = 4
Option B (e12+e22=4e_1^2 + e_2^2 = 4) is correct because the eccentricities of the hyperbola and its conjugate are both 2\sqrt{2}, and the sum of their squares is 4.
57 Medium
57 of 118
Application +4 | -1
If the normals drawn at four concyclic points on a rectangular hyperbola xy=c2x y=c^2 meet at point (α,β)(\alpha, \beta) then the centre of the circle has the coordinates
View answer and explanation
Correct option: C
(α2,β2)\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)
Option C ((α2,β2)\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)) is correct because the center of the circle is the midpoint of the line joining the point of concurrency of normals and the center of the hyperbola.
58 Medium
58 of 118
Application +4 | -1
The product of lengths of perpendicular from any point on the hyperbola x2−y2=8{x^2} - {y^2} = 8 to its asymptotes is
View answer and explanation
Correct option: D
4
Option D (4) is correct because it represents the product of the lengths of the perpendiculars from any point on the given hyperbola to its asymptotes.
59 Medium
59 of 118
Analyzing +4 | -1
Let the major axis of a standard ellipse equals the transverse axis of a standard hyperbola and their director circles have radius equal to 2R2 R and RR respectively. If e1e_1 and e2e_2 are the eccentricities of the ellipse and hyperbola, then the correct relation is
View answer and explanation
Correct option: C
4e22−e12=64 e_2^2-e_1^2=6
Option C (4e22−e12=64 e_2^2-e_1^2=6) is correct because it is derived by relating the eccentricities of the ellipse and hyperbola to the radii of their director circles.
60 Medium
60 of 118
Analyzing +4 | -1
If a circles x2+y2=a2x^2+y^2=a^2 and the rectangular hyperbola xy=c2x y=c^2 intersect in four points, (ctr,ctr)\left(c t_r, \frac{c}{t_r}\right), r=1,2,3,4r=1,2,3,4 then t1t2t3t4t_1 t_2 t_3 t_4 is equal to
View answer and explanation
Correct option: B
1
Option B is correct. Substituting the parametric equations of the hyperbola into the equation of the circle and solving for the product of the roots yields 1.
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