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JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Mixed difficulty

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121 Easy
121 of 429
Application +4 | -1
The eccentricity of the hyperbola which passes through (3, 0) and (32,2)(3\sqrt 2 ,2), is
View answer and explanation
Correct option: B
133\frac{{\sqrt {13} }}{3}
Option B 133\frac{{\sqrt {13} }}{3} is correct because it is the eccentricity calculated from the given points on the hyperbola.
122 Easy
122 of 429
Analyzing +4 | -1
If ee and e′e^{\prime} be the eccentricities of two conics S=0S=0 and S′=0S^{\prime}=0 and if e2+e′2=3e^2+e^{\prime 2}=3, then both SS and S′S^{\prime} can be
View answer and explanation
Correct option: A
Hyperbolas
Option A (Hyperbolas) is correct because for hyperbolas, the eccentricity is greater than 1, and it's possible to find two eccentricities that satisfy the given condition e2+e′2=3e^2 + e'^2 = 3.
123 Medium
123 of 429
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.
124 Easy
124 of 429
Application +4 | -1
The equation x2−4y2−2x+16y−40=0x^2-4 y^2-2 x+16 y-40=0 represents
View answer and explanation
Correct option: C
A hyperbola
Option C (A hyperbola) is correct because the equation can be rearranged into the standard form of a hyperbola equation.
125 Medium
125 of 429
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.
126 Easy
126 of 429
Application +4 | -1
The eccentricity of the hyperbola x2−y2=25x^2-y^2=25 is
View answer and explanation
Correct option: A
2\sqrt{2}
Option A (2\sqrt{2}) is correct because it is the calculated eccentricity of the given hyperbola after converting it to standard form.
127 Easy
127 of 429
Application +4 | -1
If a hyperbola passing through the origin has 3x−4y−1=03 x-4 y-1=0 and 4x−3y−6=04 x-3 y-6=0 as its asymptotes, then the equations of its transverse and conjugate axes, are
View answer and explanation
Correct option: C
x+y−5=0x+y-5=0 and x−y−1=0x-y-1=0
Option D (x+y−1=0x+y-1=0 and x−y−5=0x-y-5=0) is correct because these are the angle bisectors of the given asymptotes, representing the transverse and conjugate axes.
128 Easy
128 of 429
Application +4 | -1
The distance between the tangent to the hyperbola x24−y23=1,\frac{x^2}{4}-\frac{y^2}{3}=1, parallel to the line y=x+2y=x+2 is
View answer and explanation
Correct option: C
2\sqrt{2}
Option C (2\sqrt{2}) is correct because the distance between the tangents parallel to the given line is calculated using the formula for tangents to a hyperbola.
129 Easy
129 of 429
Application +4 | -1
The equation of the tangent to the curve 4x2−9y2=14 x^2-9 y^2=1 which is parallel to 4y=5x+74 y=5 x+7, is
View answer and explanation
Correct option: C
24y−30x=±16124 y-30 x=\pm \sqrt{161}
Option C (24y−30x=±16124 y-30 x=\pm \sqrt{161}) is correct because it represents the tangent lines to the hyperbola with the specified slope, derived using the tangent equation y=mx±a2m2−b2y = mx \pm \sqrt{a^2m^2 - b^2}.
130 Easy
130 of 429
Application +4 | -1
The coordinates of the point at which the line 3x+4y3 x+4 y =7=7 is a normal to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, are
View answer and explanation
Correct option: A
(7a23(a2+b2),7b24(a2+b2))\left(\frac{7 a^2}{3\left(a^2+b^2\right)}, \frac{7 b^2}{4\left(a^2+b^2\right)}\right)
Option A is correct because the coordinates are found by comparing the given normal equation with the standard normal equation of a hyperbola.
131 Hard
131 of 429
Analyzing +4 | -1
The normal at a point PP on the ellipse x2+4y2=16{x^2} + 4{y^2} = 16 meets the x−axisx - axis at QQ. If MM is the mid point of the line segment PQPQ, then the locus of MM intersects the latusrectum of the given ellipse at the points
View answer and explanation
Correct option: C
(±23,±17)\left( { \pm 2\sqrt 3 , \pm \frac{1}{7}} \right)
Option D ((±23,±17)\left( { \pm 2\sqrt 3 , \pm \frac{1}{7}} \right)) is correct because the locus of M intersects the latus rectum of the ellipse at these points.
132 Easy
132 of 429
Application +4 | -1
The equation of the hyperbola, the length of whose latusrectum is 8 and eccentricity is 3/53 / \sqrt{5}, is
View answer and explanation
Correct option: B
4x2−5y2=1004 x^2-5 y^2=100
Option B (4x2−5y2=1004 x^2-5 y^2=100) is correct because it correctly uses the given latus rectum and eccentricity to derive the hyperbola equation.
133 Medium
133 of 429
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.
134 Easy
134 of 429
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.
135 Easy
135 of 429
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.

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136 Easy
136 of 429
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.
137 Easy
137 of 429
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}.
138 Easy
138 of 429
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.
139 Medium
139 of 429
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.
140 Easy
140 of 429
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.
141 Easy
141 of 429
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.
142 Easy
142 of 429
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}}.
143 Easy
143 of 429
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).
144 Easy
144 of 429
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.
145 Easy
145 of 429
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.
146 Medium
146 of 429
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.
147 Easy
147 of 429
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.
148 Medium
148 of 429
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.
149 Medium
149 of 429
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.
150 Easy
150 of 429
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.
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