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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 91-120 of 429 questions

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91 Easy
91 of 429
Application +4 | -1
Tangents are drawn to the hyperbola x29−y24=1\frac{x^2}{9}-\frac{y^2}{4}=1 parallel to the straight line 2x−y=12 x-y=1. The points of contact of the tangents on the hyperbola, are
View answer and explanation
Correct option: A
(±922,±12)\left(\pm \frac{9}{2 \sqrt{2}}, \pm \frac{1}{\sqrt{2}}\right)
Option A (±922,±12)\left(\pm \frac{{9}}{{2 \sqrt{{2}}}}, \pm \frac{{1}}{{\sqrt{{2}}}}\right) is correct because it is derived from the formula for points of contact of tangents to a hyperbola.
92 Hard
92 of 429
Application +4 | -1
If the tangent at (α,β)\left( {\alpha ,\beta } \right) to the hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 cuts the auxiliary circle at points whose ordinates are y1{y_1} and y2{y_2}, then 1y1+1y2=\frac{1}{{{y_1}}} + \frac{1}{{{y_2}}} =
View answer and explanation
Correct option: D
2β\frac{2}{\beta }
Option D (2β\frac{2}{\beta }) is correct because it is derived from finding the intersection of the tangent to the hyperbola and its auxiliary circle, then using Vieta's formulas.
93 Easy
93 of 429
Application +4 | -1
The equation of the director circle of the hyperbola 9x2−16y2=1449{x^2} - 16{y^2} = 144 is
View answer and explanation
Correct option: A
x2+y2=7{x^2} + {y^2} = 7
Option A (x2+y2=7{x^2} + {y^2} = 7) is correct because the equation of the director circle is x2+y2=a2−b2{x^2} + {y^2} = a^2 - b^2 for a hyperbola.
94 Medium
94 of 429
Analyzing +4 | -1
The foci of a hyperbola are (−5,18)(-5,18) and (10,20)(10,20) and it touches the yy-axis. The length of its transverse axis, is
View answer and explanation
Correct option: C
89\sqrt{89}
Option C (89\sqrt{89}) is correct because it represents the length of the transverse axis of the hyperbola, calculated using the distance between the foci and the condition that the hyperbola touches the y-axis.
95 Easy
95 of 429
Application +4 | -1
The equation of the hyperbola whose foci are (6,5),(−4,5)\left( {6,5} \right),\left( { - 4,5} \right) and eccentricity 54\frac{5}{4}, is
View answer and explanation
Correct option: A
(x−1)216−(y−5)29=1\frac{{{{\left( {x - 1} \right)}^2}}}{{16}} - \frac{{{{\left( {y - 5} \right)}^2}}}{9} = 1
Option A ((x−1)216−(y−5)29=1\frac{{{{\left( {x - 1} \right)}^2}}}{{16}} - \frac{{{{\left( {y - 5} \right)}^2}}}{9} = 1) is correct because it represents the equation of a hyperbola with the given foci and eccentricity, derived by finding the center, a, and b.
96 Easy
96 of 429
Application +4 | -1
For the ellipse 3x2+4y2+6x−8y−5=03{x^2} + 4{y^2} + 6x - 8y - 5 = 0 the eccentricity is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because after completing the squares and transforming the equation into standard form, the eccentricity of the ellipse is calculated using the formula e=1−b2a2e = \sqrt{1 - \frac{b^2}{a^2}}.
97 Easy
97 of 429
Application +4 | -1
Given the points A(0,4)A\left( {0,4} \right) and B(0,−4)B\left( {0, - 4} \right), then the equation of the locus of the point P(x,y)P\left( {x,y} \right) such that, ∣AP−BP∣=6\left| {AP - BP} \right| = 6, is
View answer and explanation
Correct option: D
y29−x27=1\frac{{{y^2}}}{9} - \frac{{{x^2}}}{7} = 1
Option D (y29−x27=1\frac{{{y^2}}}{9} - \frac{{{x^2}}}{7} = 1) is correct because it represents the equation of a hyperbola derived from the given condition ∣AP−BP∣=6\left| {AP - BP} \right| = 6.
98 Easy
98 of 429
Application +4 | -1
If the line x−1=0x-1=0 is a directrix of the hyperbola kx2−y2=6k x^2-y^2=6, then the hyperbola passes through the point
View answer and explanation
Correct option: C
(5,−2)(\sqrt{5},-2)
Option C is correct because by finding the eccentricity and using the directrix equation, we can determine the value of k and then check which point satisfies the hyperbola equation.
99 Easy
99 of 429
Application +4 | -1
The line 93x+12y=23439 \sqrt{3} x+12 y=234 \sqrt{3} is a normal to the hyperbola x281−y236=1\frac{x^2}{81}-\frac{y^2}{36}=1 at the points.
View answer and explanation
Correct option: A
(18,63)(18,6 \sqrt{3})
Option A ((18,63)(18, 6\sqrt{3})) is correct because the point satisfies the equation of the normal to the hyperbola at that point.
100 Easy
100 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 the point (8,33)(8,3 \sqrt{3}) is
View answer and explanation
Correct option: D
2x+3y=252 x+\sqrt{3} y=25
Option D (2x+3y=252 x+\sqrt{3} y=25) is correct because substituting the point (8, 3√3) into the equation satisfies the equation, and it represents the normal to the hyperbola at that point.
101 Easy
101 of 429
Application +4 | -1
The product of the lengths of perpendiculars drawn from any point on the hyperbola x2−2y2−2=0{x^2} - 2{y^2} - 2 = 0 to its asymptotes is
View answer and explanation
Correct option: B
23\frac{2}{3}
Option B (23\frac{2}{3}) is correct because the product of perpendiculars from any point on the hyperbola to its asymptotes is a constant value determined by the hyperbola's parameters.
102 Hard
102 of 429
Analyzing +4 | -1
The normal at PP to a hyperbola of eccentricity ee, intersects its transverse and conjugate axes at LL and MM respectively. If locus of the mid point of LMLM is hyperbola, then eccentricity of the hyperbola is
View answer and explanation
Correct option: B
e(e2−1)\frac{e}{{\sqrt {\left( {{e^2} - 1} \right)} }}
Option B (\frac{e}{{\sqrt {\left( {{{e^2} - 1} \right)} }}\frac{e}{{\sqrt {\left( {{{e^2} - 1} \right)} }}) is correct because it is derived from the geometry of the hyperbola and the properties of the normal line intersecting the transverse and conjugate axes.
103 Easy
103 of 429
Application +4 | -1
If lx+my+n=0lx + my + n = 0 is a tangent to the rectangular hyperbola xy=c2,xy = {c^2}, then
View answer and explanation
Correct option: A
l<m<0l < m < 0
Option D (None of these) is correct because the condition for tangency to a rectangular hyperbola leads to a different relationship between l, m, and n.
104 Easy
104 of 429
Application +4 | -1
Focus of hyperbola is (±3,0)\left( { \pm 3,0} \right) and equation of tangent is 2x+y−4=02x + y - 4 = 0, find the equation of hyperbola
View answer and explanation
Correct option: A
4x2−5y2=204{x^2} - 5{y^2} = 20
Option A (4x2−5y2=204{x^2} - 5{y^2} = 20) is correct because it satisfies the given conditions for the foci and the tangent.
105 Easy
105 of 429
Application +4 | -1
If length of the transverse axis of a hyperbola is 8 and its eccentricity is 5/2\sqrt{5} / 2 then the length of a latus rectum of the hyperbola is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the length of the latus rectum is calculated using the formula 2b2a\frac{{2b^2}}{{a}} and the given information.
106 Easy
106 of 429
Application +4 | -1
The eccentricity of the hyperbola in the standard form x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1, passing through (3,0) and (3,2,2)\left( {3,\sqrt 2 ,2} \right) is
View answer and explanation
Correct option: D
133\frac{{\sqrt {13} }}{3}
Option D 133\frac{{\sqrt {13} }}{3} is correct because it follows from substituting the given points into the hyperbola equation and solving for eccentricity using the formula e=1+b2a2e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}}.
107 Easy
107 of 429
Application +4 | -1
The eccentricity of the hyperbola x2−y2=2004{x^2} - {y^2} = 2004 is
View answer and explanation
Correct option: D
2\sqrt 2
Option D (2\sqrt 2) is correct because the given hyperbola is a rectangular hyperbola, and the eccentricity of a rectangular hyperbola is always 2\sqrt{2}.
108 Easy
108 of 429
Application +4 | -1
If the line 2x+6y=22x + \sqrt 6 y = 2 touches the hyperbola x2−2y2=4{x^2} - 2{y^2} = 4, then the point of contact is
View answer and explanation
Correct option: D
(4,−6)(4, - \sqrt {6)}
Option D ((4,−6)(4, - \sqrt {6)}) is correct because the point of contact is found by comparing the tangent equation to the hyperbola with the given line equation.
109 Medium
109 of 429
Application +4 | -1
The equation of the hyperbola whose vertices are (±2,0)( \pm 2,0) and foci are (±3,0)( \pm 3,0) is x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. Sum of a2{a}^2 and b2{b}^2 is
View answer and explanation
Correct option: A
9
Option A (9) is correct because the sum of a² and b² for the given hyperbola is 9, calculated using the vertices and foci.
110 Medium
110 of 429
Application +4 | -1
PP and QQ are two points on the rectangular hyperbola xy=c2x y=c^2 such that the abscissa of PP and QQ are the roots of the equations x2−6x−16=0x^2-6 x-16=0. Equation of the chord joining PP and QQ is
View answer and explanation
Correct option: C
c2x−16y=6c2c^2 x-16 y=6 c^2
Option C (c2x−16y=6c2c^2 x-16 y=6 c^2) is correct because it is the equation of the chord joining P and Q on the rectangular hyperbola.
111 Easy
111 of 429
Application +4 | -1
The angle between the asymptotes of the hyperbola 27x2−9y2=2427{x^2} - 9{y^2} = 24 is
View answer and explanation
Correct option: B
1200{120^0}
Option B (1200{120^0}) is correct because it is calculated using the formula 2 * arctan(b/a) after rewriting the hyperbola equation in standard form.
112 Medium
112 of 429
Application +4 | -1
Two tangents are drawn to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 such that the product of their slopes is c2c^2. The locus of their point of intersection is
View answer and explanation
Correct option: C
y2+b2=c2(x2−a2)y^2+b^2=c^2\left(x^2-a^2\right)
Option C (y2+b2=c2(x2−a2)y^2+b^2=c^2\left(x^2-a^2\right)) is correct because the locus of the point of intersection of two tangents with a product of slopes equal to c2c^2 is given by this equation.
113 Medium
113 of 429
Analyzing +4 | -1
If the tangent at a point PP on x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 cuts one of its directrices at QQ, then the angle made by PQPQ at the corresponding focus, is
View answer and explanation
Correct option: D
900{90^0}
Option D (900{90^0}) is correct because the tangent at a point P on the hyperbola cuts the directrix at Q, and PQ subtends a right angle at the corresponding focus.
114 Easy
114 of 429
Application +4 | -1
The eccentricity of the hyperbola 19993(x2−y2)=1,\frac{{\sqrt {1999} }}{3}\left( {{x^2} - {y^2}} \right) = 1, is
View answer and explanation
Correct option: A
2\sqrt 2
Option A (2\sqrt 2) is correct because the given hyperbola is a rectangular hyperbola, and rectangular hyperbolas always have an eccentricity of 2\sqrt 2.
115 Medium
115 of 429
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}.
116 Medium
116 of 429
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.
117 Medium
117 of 429
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}.
118 Medium
118 of 429
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.
119 Easy
119 of 429
Application +4 | -1
If two perpendicular tangents are drawn from a point (α,β)(\alpha, \beta) to the hyperbola x2−y2=16x^2-y^2=16, then the locus of (α,β)(\alpha, \beta) is
View answer and explanation
Correct option: A
A pair of straight line
The correct answer is A (A pair of straight line) because the locus of the point from which perpendicular tangents are drawn to the hyperbola is a pair of straight lines.
120 Easy
120 of 429
Application +4 | -1
Let the eccentricity of the hyperbola H:x2a2−y2b2=1H:\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 be 52\sqrt {\frac{5}{2}} and length of its latus rectum be 62.6\sqrt 2 . If y=2x+cy = 2x + c is a tangent to the hyperbola H,H, Then the value of c2{c^2} is equal to
View answer and explanation
Correct option: B
20
Option B (20) is correct because the value of c2 is calculated using the eccentricity and latus rectum to find a and b, then applying the tangent condition.
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