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JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Easy

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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 298 questions

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91 Easy
91 of 298
Application +4 | -1
If the eccentricity of the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 is 5\sqrt{5} and the distance between the foci is 12, then b2−a2b^2-a^2 is equal to (3/5)k2(3 / 5) k^2 where kk is equal to
View answer and explanation
Correct option: D
6
Option D (6) is correct because by using the given eccentricity and distance between foci, we can find the value of a2a^2 and then calculate b2−a2b^2-a^2 which leads to k=6.
92 Easy
92 of 298
Application +4 | -1
The length of the chord of the parabola y2=4axy^2=4 a x, which passes through the vertex and makes an angle θ\theta with the axis of the parabola, is
View answer and explanation
Correct option: A
4acos⁡θcosec⁡2θ4 a \cos \theta \operatorname{cosec}^2 \theta
Option A (4acos⁡θcosec⁡2θ4 a \cos \theta \operatorname{{cosec}}^2 \theta) is correct because it represents the length of the chord calculated using the intersection points of the parabola and the line.
93 Easy
93 of 298
Application +4 | -1
The equation of a hyperbola, whose foci are (5,0)(5,0) and (−5,0)(-5,0) and the length of whose conjugate axis is 8 , is
View answer and explanation
Correct option: B
16x2−9y2=14416 x^2-9 y^2=144
Option B (16x2−9y2=14416 x^2-9 y^2=144) is correct because given the foci and conjugate axis length, we can determine the values of 'a' and 'b' and thus construct the hyperbola's equation.
94 Easy
94 of 298
Application +4 | -1
The tangent at point (2sec⁡θ,3tan⁡θ)(2 \sec \theta, 3 \tan \theta) of the 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 θ\theta is
View answer and explanation
Correct option: A
30∘30^{\circ}
Option A (30∘30^{\circ}) is correct because the slope of the tangent to the hyperbola at the given point must equal the slope of the parallel line, leading to the equation sec⁡(θ)=2tan⁡(θ)\sec(\theta) = 2 \tan(\theta), which simplifies to sin⁡θ=12\sin \theta = \frac{1}{2}, thus θ=30∘\theta = 30^{\circ}.
95 Easy
95 of 298
Application +4 | -1
If e1e_1 is the eccentricity of the hyperbola x249−y236=1\frac{x^2}{49}-\frac{y^2}{36}=1 and e2e_2 is the eccentricity of the hyperbola x236−y249=1\frac{x^2}{36}-\frac{y^2}{49}=1, then
View answer and explanation
Correct option: C
1e12+1e22=1\frac{1}{e_1^2}+\frac{1}{e_2^2}=1
Option C (1e12+1e22=1\frac{1}{e_1^2}+\frac{1}{e_2^2}=1) is correct because the eccentricities of the two hyperbolas are related by this equation.
96 Easy
96 of 298
Application +4 | -1
The normal to the hyperbola 4x2−9y2=364 x^2-9 y^2=36 meets the axes in M and N{N} and the lines MP,NPMP, NP are drawn at right angles to the axes. The locus of P{P} is the hyperbola
View answer and explanation
Correct option: A
9x2−4y2=1699 x^2-4 y^2=169
Option A is correct because the locus of P is a hyperbola, derived from the normal equation and the given conditions.
97 Easy
97 of 298
Understanding +4 | -1
The equation x212−k+y28−k=1\frac{{{x^2}}}{{12 - k}} + \frac{{{y^2}}}{{8 - k}} = 1 represents
View answer and explanation
Correct option: C
A hyperbola if 8<k<128 < k < 12
Option C (A hyperbola if 8 < k < 12) is correct because for the equation to represent a hyperbola, the denominators must have opposite signs, implying 8 < k < 12.
98 Easy
98 of 298
Application +4 | -1
Distance between the directrices of the hyperbola x249−y216=1\frac{x^2}{49}-\frac{y^2}{16}=1 is
View answer and explanation
Correct option: D
9865\frac{98}{\sqrt{65}}
Option D (9865\frac{{98}}{{\sqrt{{65}}}}) is correct because the distance between the directrices is calculated using the formula 2ae\frac{{2a}}{{e}}.
99 Easy
99 of 298
Understanding +4 | -1
The difference in focal distance of any point on the hyperbola x216−y29=1\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1 is
View answer and explanation
Correct option: A
8
Option A (8) is correct because the difference in focal distances of any point on a hyperbola is equal to the length of its transverse axis (2a).
100 Easy
100 of 298
Application +4 | -1
If the line y=3x+λy = 3x + \lambda touches the hyperbola 9x2−5y2=45,9{x^2} - 5{y^2} = 45, then the value of λ\lambda is
View answer and explanation
Correct option: C
6
Option C (6) is correct because it is the value of λ that satisfies the condition for the line to be tangent to the hyperbola.
101 Easy
101 of 298
Application +4 | -1
The eccentricity of the hyperbola 9x2−16y2−18x−64y−199=09{x^2} - 16{y^2} - 18x - 64y - 199 = 0 is
View answer and explanation
Correct option: B
54\frac{5}{4}
Option B (54\frac{5}{4}) is correct because it is the calculated eccentricity of the given hyperbola after rewriting it in standard form.
102 Easy
102 of 298
Application +4 | -1
The foci of the hyperbola 2x2−3y2=52{x^2} - 3{y^2} = 5 are
View answer and explanation
Correct option: A
(±56,0)\left( { \pm \frac{5}{{\sqrt 6 }},0} \right)
Option A ((±56,0)\left( { \pm \frac{5}{{\sqrt 6 }},0} \right)) is correct because it represents the foci of the given hyperbola, calculated using the eccentricity and semi-major axis.
103 Easy
103 of 298
Analyzing +4 | -1
If the sum of the slopes of the normals from a point PP to the hyperbola xy=c2x y=c^2 is equal to λ(λ∈R+)\lambda\left(\lambda \in R^{+}\right), then locus of point PP is
View answer and explanation
Correct option: D
x2=λc2x^2=\lambda c^2
Option D (x2=λc2x^2=\lambda c^2) is correct because it represents the locus of point PP derived from the condition that the sum of the slopes of the normals from PP to the hyperbola is equal to λ\lambda.
104 Easy
104 of 298
Application +4 | -1
The line 5x+12y=95x + 12y = 9 touches the hyperbola x2−9y2=9{x^2} - 9{y^2} = 9 at the point
View answer and explanation
Correct option: B
(5,−4/3)\left( {5, - 4/3} \right)
Option B ((5,−4/3)\left( {5, - 4/3} \right)) is correct because it is the point that satisfies both the equation of the hyperbola and the equation of the tangent line.
105 Easy
105 of 298
Analyzing +4 | -1
If f(x)=ax3+bx2+cx+d,(a,b,c,df(x)=a x^3+b x^2+c x+d,(a, b, c, d are rational numbers) and roots of f(x)=0f(x)=0 are eccentricities of a parabola and a rectangular hyperbola, then a+b+c+da+b+c+d equals
View answer and explanation
Correct option: B
0
Option B (0) is correct because the roots of f(x)=0f(x)=0 are 1, 2\sqrt{2}, and −2-\sqrt{2}, and substituting these into the polynomial gives a+b+c+d=0a+b+c+d = 0.
106 Easy
106 of 298
Application +4 | -1
Coordinates of foci of hyperbola are (−5,3)\left( { - 5,3} \right) and (7,3)\left( {7,3} \right) and eccentricity is 32\frac{3}{2}. Then ,length of its latusrectum is
View answer and explanation
Correct option: B
10
Option B (10) is correct because the length of the latus rectum is calculated using the formula 2b2a\frac{2b^2}{a} after finding the values of a and b from the given foci and eccentricity.
107 Easy
107 of 298
Application +4 | -1
Eccentricity of hyperbola whose asymptotes are 3x−4y=73x - 4y = 7 and 4x+3y=84x + 3y = 8, is
View answer and explanation
Correct option: A
2\sqrt 2
Option A (2\sqrt 2) is correct because the given asymptotes are perpendicular, indicating a rectangular hyperbola with eccentricity 2\sqrt 2.
108 Easy
108 of 298
Analyzing +4 | -1
Consider the hyperbola H:x2−y2=1H: x^2-y^2=1 and a circle S with centre N(x2,0)N\left(x_2, 0\right). Suppose that HH and SS touch each other at a point P(x1,y1)P\left(x_1, y_1\right) with x1>1x_1>1 and y1>0y_1>0. The common tangent to HH and SS at PP intersects the xx-axis at point MM. If (l,m)(l, m) is the centroid of triangle ΔPMN\Delta P M N, then the incorrect expression, is
View answer and explanation
Correct option: C
dldx1=1+13x12\frac{d l}{d x_1}=1+\frac{1}{3 x_1{ }^2} for x1>1x_1>1
The incorrect expression is C. The correct derivative of l with respect to x1 is 1 - 1/(3x12), not 1 + 1/(3x12).
109 Easy
109 of 298
Application +4 | -1
If the distance between the directrices of a rectangular hyperbola is 10 unit, then the distance between its foci is
View answer and explanation
Correct option: D
20
Option D (20) is correct because the distance between the foci of a rectangular hyperbola is calculated using the given distance between the directrices and the eccentricity.
110 Easy
110 of 298
Application +4 | -1
If a hyperbola passes through the point P(2,3)P\left( {\sqrt 2 ,\sqrt 3 } \right) and has foci at (± 2,0)\left( { \pm \,2,0} \right), then the tangent to this hyperbola at PP also passes through the point
View answer and explanation
Correct option: B
(22,33)\left( {2\sqrt 2 ,3\sqrt 3 } \right)
Option B (22,33)\left( {2\sqrt 2 ,3\sqrt 3 } \right) is correct because after finding the equation of the hyperbola, we can find the equation of the tangent at the given point and check which point satisfies the tangent equation.
111 Easy
111 of 298
Application +4 | -1
If x−5y+c=0x-\sqrt{5} y+c=0 is a tangent to the hyperbola x225−y24=1\frac{x^2}{25}-\frac{y^2}{4}=1, then the value of cc is
View answer and explanation
Correct option: B
±5\pm \sqrt{5}
Option B (±5\pm \sqrt{5}) is correct because substituting the tangent equation into the hyperbola equation and solving for c yields ±5\pm \sqrt{5}.
112 Easy
112 of 298
Application +4 | -1
The latus rectum of the hyperbola 16x2−9y2=14416 x^2-9 y^2=144 is
View answer and explanation
Correct option: B
32/332 / 3
Option B (32/332 / 3) is correct because it is the latus rectum calculated using the formula 2b2/a2b^2/a for the given hyperbola.
113 Easy
113 of 298
Remembering +4 | -1
The difference of the focal distances of any point on the hyperbola is equal to its
View answer and explanation
Correct option: C
Transverse axis
Option C (Transverse axis) is correct because the difference of the focal distances of any point on a hyperbola is equal to the length of its transverse axis.
114 Easy
114 of 298
Application +4 | -1
The locus of the midpoints of the chord of the circle, x2+y2=25{x^2} + {y^2} = 25 which is tangent to the hyperbola, x29−y216=1\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{16}} = 1 is:
View answer and explanation
Correct option: D
(x2+y2)2−9x2+16y2=0{\left( {{x^2} + {y^2}} \right)^2} - 9{x^2} + 16{y^2} = 0
Option D ((x2+y2)2−9x2+16y2=0{\left( {{x^2} + {y^2}} \right)^2} - 9{x^2} + 16{y^2} = 0) is correct because it represents the locus of the midpoints of the chord of the circle that is tangent to the hyperbola, derived by equating the equation of the tangent to the hyperbola with the equation of the chord of the circle.
115 Easy
115 of 298
Application +4 | -1
The equation of the chord joining two points (x1,y1)\left(x_1, y_1\right) and (x2,y2)\left(x_2, y_2\right) on the rectangular hyperbola xy=c2x y=c^2 is
View answer and explanation
Correct option: A
xx1+x2+yy1+y2=1\frac{x}{x_1+x_2}+\frac{y}{y_1+y_2}=1
Option C (xy1+y2+yx1+x2=1\frac{x}{y_1+y_2}+\frac{y}{x_1+x_2}=1) is correct because it represents the equation of the chord joining two points on the rectangular hyperbola, derived using the midpoint formula and the T = S' method.
116 Easy
116 of 298
Remembering +4 | -1
The eccentricity ee of the hyperbola −x2a2+y2b2=1-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is given by
View answer and explanation
Correct option: D
a2+b2b2\sqrt{\frac{a^2+b^2}{b^2}}
Option D (a2+b2b2\sqrt{\frac{{a^2+b^2}}{{b^2}}}) is correct because the eccentricity of the hyperbola is calculated using the formula e=1+a2b2e = \sqrt{1 + \frac{a^2}{b^2}}.
117 Easy
117 of 298
Application +4 | -1
The equation of the chord of the hyperbola 4x2−y2=4 x^2-y^2= 4 , which is bisected at the point (2,−3)(2,-3) is
View answer and explanation
Correct option: A
8x+3y=78 x+3 y=7
Option A is correct because the equation of the chord bisected at a point is found using T = S1.
118 Easy
118 of 298
Application +4 | -1
The length of the transverse axis of the rectangular hyperbola xy=18xy = 18 is
View answer and explanation
Correct option: B
12
Option B (12) is correct because the length of the transverse axis of the rectangular hyperbola is 2a, where a is calculated from the equation xy = a²/2.
119 Easy
119 of 298
Application +4 | -1
The value of mm, for which the line y=mx+2  y = mx + 2\; becomes a tangent to the conic 4x2−9y2=364{x^2} - 9{y^2} = 36 are
View answer and explanation
Correct option: B
±223\pm \frac{{2\sqrt 2 }}{3}
Option B (±223\pm \frac{{2\sqrt 2 }}{3}) is correct because these are the values of m for which the line y = mx + 2 is tangent to the hyperbola.
120 Easy
120 of 298
Application +4 | -1
If the latusrectum subtends a right angle at the centre of the hyperbola x2a2−y2b2=1,\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1, then its eccentricity is
View answer and explanation
Correct option: C
5+12\frac{{\sqrt 5 + 1}}{2}
Option C (5+12\frac{{\sqrt 5 + 1}}{2}) is correct because the eccentricity is calculated using the condition that the latus rectum subtends a right angle at the center of the hyperbola.
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