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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 151-180 of 298 questions

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151 Easy
151 of 298
Application +4 | -1
The locus of the middle points of chords of hyperbola 3x2−2y2+4x−6y=03 x^2-2 y^2+4 x-6 y=0 parallel to y=2xy=2 x is
View answer and explanation
Correct option: A
3x−4y=43 x-4 y=4
Option A (3x−4y=43 x-4 y=4) is correct because the locus of the midpoints of chords parallel to a given line can be found using the T = S1 method, and then substituting the slope condition.
152 Easy
152 of 298
Application +4 | -1
A common tangent to the conics x2=6yx^{2}=6 y and 2x2−4y2=92 x^{2}-4 y^{2}=9 is:
View answer and explanation
Correct option: A
x−y=32x-y=\frac{3}{2}
Option A (x−y=32x-y=\frac{3}{2}) is correct because it represents a common tangent to both the given conics.
153 Easy
153 of 298
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. Find the equation of the hyperbola, if its eccentricity is 2, is
View answer and explanation
Correct option: A
x24−y212=1\frac{x^2}{4}-\frac{y^2}{12}=1
Option A (x24−y212=1\frac{x^2}{4}-\frac{y^2}{12}=1) is correct because it correctly calculates the foci of the ellipse and uses them to find the equation of the hyperbola with the given eccentricity.
154 Easy
154 of 298
Application +4 | -1
If the tangent and normal to a rectangular hyperbola cut off intercepts a1a_1 and a2a_2 on one axis and b1b_1 and b2b_2 on the other, then
View answer and explanation
Correct option: A
a1a2+b1b2=0a_1 a_2+b_1 b_2=0
Option A (a1a2+b1b2=0a_1 a_2+b_1 b_2=0) is correct because the product of the intercepts of the tangent and normal on one axis plus the product of the intercepts on the other axis equals zero for a rectangular hyperbola.
155 Easy
155 of 298
Application +4 | -1
If the latus rectum of a hyperbola be 8 and its eccentricity be 3/53 / \sqrt{5}, then the equation of the hyperbola is
View answer and explanation
Correct option: A
4x2−5y2=1004 x^2-5 y^2=100
Option A (4x2−5y2=1004 x^2-5 y^2=100) is correct because it satisfies the given conditions for the latus rectum and eccentricity of the hyperbola.
156 Easy
156 of 298
Analyzing +4 | -1
The one which does not represent a hyperbola is
View answer and explanation
Correct option: D
x2−y2=0{x^2} - {y^2} = 0
Option D (x2−y2=0{x^2} - {y^2} = 0) is correct because it represents a pair of straight lines, not a hyperbola.
157 Easy
157 of 298
Application +4 | -1
The mid-point of the chord intercepted by the hyperbola 9x2−16y2=1449{x^2} - 16{y^2} = 144 on the line 9x−8y−10=0,9x - 8y - 10 = 0, is
View answer and explanation
Correct option: D
(2,1)\left( {2,1} \right)
Option D ((2,1)) is correct because it is found by using the T=S' method to find the midpoint of the chord intercepted by the line and hyperbola.
158 Easy
158 of 298
Application +4 | -1
If the eccentricities of the hyperbolas x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 and y2b2−x2a2=1\frac{y^2}{b^2}-\frac{x^2}{a^2}=1 be ee and e1e_1, then find the value of 1e2+1e12\frac{1}{e^2}+\frac{1}{e_1^2}.
View answer and explanation
Correct option: A
1
Option A (1) is correct because the sum of the reciprocals of the squares of the eccentricities of the two hyperbolas is equal to 1.
159 Easy
159 of 298
Application +4 | -1
If 2x−y+1=02 x-y+1=0 is a tangent to hyperbola x2a2+y216=1\frac{x^2}{a^2}+\frac{y^2}{16}=1, then which of the following are sides of a right angled triangle?
View answer and explanation
Correct option: A
2a,4,12 a, 4,1
Option A (2a,4,12a, 4, 1) is correct because the tangent condition to the hyperbola leads to a relationship between aa and the constant term, resulting in these side lengths forming a right-angled triangle.
160 Easy
160 of 298
Application +4 | -1
The equation of the hyperbola in the standard form (with transverse axis along the xx-axis) having the length of the latusrectum =9= 9 unit and eccentricity =54= \frac{5}{4}, is
View answer and explanation
Correct option: C
x264−y236=1\frac{{{x^2}}}{{64}} - \frac{{{y^2}}}{{36}} = 1
Option C (x264−y236=1\frac{{{x^2}}}{{64}} - \frac{{{y^2}}}{{36}} = 1) is correct because it satisfies the given latus rectum length and eccentricity for a hyperbola.
161 Easy
161 of 298
Application +4 | -1
Let a>0,b>0a>0, b>0. Let ee and ll respectively be the eccentricity and length of the latus rectum of the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. Let e′e^{\prime} and l′l^{\prime} respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If e2=1114l\mathrm{e}^2=\frac{11}{14} l and (e′)2=118l′\left(\mathrm{e}^{\prime}\right)^2=\frac{11}{8} l^{\prime}, then the value of 77a+44b77 a+44 b is equal to :
View answer and explanation
Correct option: D
130
Option D (130) is correct because it is the result of solving the equations derived from the properties of the hyperbola and its conjugate.
162 Easy
162 of 298
Application +4 | -1
If ee is the eccentricity of the hyperbola x218−y26=1\frac{x^2}{18}-\frac{y^2}{6}=1 and rr is the radius of the circle x2+y2−6x−18y+87=0,x^2 + y^2 - 6x - 18y + 87 = 0, then the value of erer is equal to:
View answer and explanation
Correct option: C
2
Option C (2) is correct because it involves calculating the eccentricity of the hyperbola and the radius of the circle, then multiplying them.
163 Easy
163 of 298
Application +4 | -1
The point on the curve 3x2−4y2=723{x^2} - 4{y^2} = 72, which is nearest to the line 3x+2y−1=03x + 2y - 1 = 0, is
View answer and explanation
Correct option: B
(6,−3)\left( {6, - 3} \right)
Option B ((6,−3)\left( {6, - 3} \right)) is correct because it's the point on the curve closest to the line, determined by minimizing the distance between the point and the line.
164 Easy
164 of 298
Application +4 | -1
Foci of the rectangular hyperbola are (±7,0\pm 7,0), the equation of the hyperbola is
View answer and explanation
Correct option: C
2x2−2y2=492 x^2-2 y^2=49
Option C (2x2−2y2=492 x^2-2 y^2=49) is correct because it represents a rectangular hyperbola with the given foci.
165 Easy
165 of 298
Application +4 | -1
Number of points from where perpendicular tangents to the curve x216−y225=1\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{{25}} = 1 can be drawn, is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the director circle of the given hyperbola is an imaginary circle, indicating that there are no points from which perpendicular tangents can be drawn.
166 Easy
166 of 298
Remembering +4 | -1
The reciprocal of the eccentricity of rectangular hyperbola is
View answer and explanation
Correct option: C
2\sqrt{2}
Option D (1/21 / \sqrt{2}) is incorrect. The eccentricity of a rectangular hyperbola is 2\sqrt{2}, so the reciprocal is 1/21/\sqrt{2}.
167 Easy
167 of 298
Application +4 | -1
If e1,e2e_1, e_2 are the eccentricities of the hyperbola 2x2−2y2=12 x^2-2 y^2=1 and the ellipse x2+2y2=2x^2+2 y^2=2 respectively. Then
View answer and explanation
Correct option: B
e1e2=1e_1 e_2=1
Option B (e1e2=1e_1 e_2=1) is correct because when calculating the eccentricities of the given hyperbola and ellipse, their product equals 1.
168 Easy
168 of 298
Application +4 | -1
The length of the transverse axis of a hyperbola is 7 and it passes through the point (5,−2)(5,-2). The equation of the hyperbola is
View answer and explanation
Correct option: C
449x2−51196y2=1\frac{4}{49} x^2-\frac{51}{196} y^2=1
Option C (449x2−51196y2=1\frac{4}{49} x^2-\frac{51}{196} y^2=1) is correct because it represents a hyperbola with a transverse axis of 7 and passes through the point (5,-2).
169 Easy
169 of 298
Application +4 | -1
If the normal at PP on the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 meets the transverse axis at G,SG, S is a foci and ee the eccentricity of the hyperbola then SG:SPS G: S P is equal to
View answer and explanation
Correct option: C
ee
Option C (ee) is correct because the ratio SG:SP is equal to the eccentricity of the hyperbola.
170 Easy
170 of 298
Remembering +4 | -1
The equation ax2+2hxy+by2+2gx+2fy+c=0a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 represents a rectangular hyperbola if
View answer and explanation
Correct option: A
Δ≠0,h2>ab,a+b=0\Delta \neq 0, h^2 > a b, a+b=0
Option A (Δ≠0,h2>ab,a+b=0\Delta \neq 0, h^2 > a b, a+b=0) is correct because these are the conditions for the general equation to represent a rectangular hyperbola.
171 Easy
171 of 298
Application +4 | -1
The hyperbola x216−y29=1\frac{x^2}{16}-\frac{y^2}{9}=1 and point P(4,3)P(4,3), then find number of tangents from PP.
View answer and explanation
Correct option: C
2
Option C (2) is correct because substituting the point P(4,3) into the hyperbola's equation results in a negative value, indicating that the point lies outside the hyperbola, thus allowing for two tangents to be drawn.
172 Easy
172 of 298
Application +4 | -1
The equation of the tangent to the conic x2−y2−8x+2y+11=0{x^2} - {y^2} - 8x + 2y + 11 = 0 at (2,1)\left( {2,1} \right) is
View answer and explanation
Correct option: D
x−2=0x - 2 = 0
Option D (x−2=0x - 2 = 0) is correct because it represents the tangent line to the conic at the given point, found by using the tangent equation formula.
173 Easy
173 of 298
Application +4 | -1
The equation of a tangent to the hyperbola 16x2−25y2−96x+100y−356=016 x^2-25 y^2-96 x+100 y-356=0, which makes an angle π/4\pi / 4 with the transverse axis, is
View answer and explanation
Correct option: A
y=x+2y=x+2
Option A (y=x+2y=x+2) is correct because it represents a tangent line to the hyperbola with a slope of 1, derived using the tangent equation y−k=m(x−h)±a2m2−b2y-k=m(x-h) \pm \sqrt{a^2 m^2-b^2}.
174 Easy
174 of 298
Application +4 | -1
What is the slope of the tangent drawn to the hyperbola xy=a,  (a≠0)xy = a,\;\left( {a \ne 0} \right) at the point (a,  1)?\left( {a,\;1} \right)?
View answer and explanation
Correct option: B
−1a- \frac{1}{a}
The correct answer is B (−1a- \frac{1}{a}) because it represents the slope of the tangent to the hyperbola at the given point, calculated using differentiation.
175 Easy
175 of 298
Understanding +4 | -1
The difference of the focal distance of any point on the hyperbola 9x2−16y2=1449 x^2-16 y^2=144 is
View answer and explanation
Correct option: A
8
Option A (8) is correct because the difference of the focal distance of any point on the hyperbola is equal to 2a, where a is the semi-major axis, which is 4 in this case.
176 Easy
176 of 298
Application +4 | -1
The equation x22−λ−−y2λ−5−1=0\frac{{{x^2}}}{{2 - \lambda }}--\frac{{{y^2}}}{{\lambda - 5}} - 1 = 0, represent an ellipse, if
View answer and explanation
Correct option: B
λ<2\lambda < 2
Option B (λ<2\lambda < 2) is correct because for the equation to represent an ellipse, both denominators must be positive, leading to the condition λ<2\lambda < 2.
177 Easy
177 of 298
Application +4 | -1
If m1m_1 and m2m_2 are two values of mm for which the line y=mx+25y=m x+2 \sqrt{5} is a tangent to the hyperbola x24−y216=1\frac{x^2}{4}-\frac{y^2}{16}=1, then the value of ∣m1+1m2∣\left|m_1+\frac{1}{m_2}\right| is equal to
View answer and explanation
Correct option: A
83\frac{8}{3}
Option A (83\frac{{8}}{{3}}) is correct because it involves finding the values of m for which the line is tangent to the hyperbola and then calculating the absolute value of the expression.
178 Easy
178 of 298
Application +4 | -1
The equation of the hyperbola whose foci are (6, 4) and (−4,4)( - 4,4) and eccentricity 2, is
View answer and explanation
Correct option: D
4(x−1)225−4(y−4)275=1\frac{{4{{\left( {x - 1} \right)}^2}}}{{25}} - \frac{{4{{\left( {y - 4} \right)}^2}}}{{75}} = 1
Option D (4(x−1)225−4(y−4)275=1\frac{{4{{\left( {x - 1} \right)}^2}}}{{25}} - \frac{{4{{\left( {y - 4} \right)}^2}}}{{75}} = 1) is correct because it satisfies the given foci and eccentricity for a hyperbola centered at (1, 4).
179 Easy
179 of 298
Application +4 | -1
If the focii of the 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 coincides then k=k=?
View answer and explanation
Correct option: A
±3\pm \sqrt{3}
Option A (±3\pm \sqrt{3}) is correct because equating the focal distances of the ellipse and hyperbola leads to the equation k2a2−a2=a2+a2k^2 a^2 - a^2 = a^2 + a^2, which simplifies to k=±3k = \pm \sqrt{3}.
180 Easy
180 of 298
Application +4 | -1
The pole of the line lx+my+n=0l x+m y+n=0 with respect to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, is
View answer and explanation
Correct option: B
(−a2ln,b2mn)\left(-\frac{a^2 l}{n}, \frac{b^2 m}{n}\right)
Option C ((a2ln,−b2mn)\left(\frac{{a^2 l}}{n},-\frac{{b^2 m}}{n}\right)) is correct because the pole of a line with respect to a hyperbola can be found by comparing the equation of the polar with the given line.
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