Skip to main content
Back to HYPERBOLA

Your Progress

0 / 118 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Medium

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 91-118 of 118 questions

Preparing practice and saved progress…

91 Medium
91 of 118
Application +4 | -1
If the equation of the hyperbola is y29−x227=1\frac{y^2}{9}-\frac{x^2}{27}=1, then

I. the coordinates of the foci are (0,±6)(0, \pm 6)

II. the length of the latus rectum is 18 units

III. the eccentricity is 45\frac{4}{5}
View answer and explanation
Correct option: C
Only I and II is true
Option C is correct because statements I and II are true. The foci and latus rectum are correctly calculated, while the eccentricity is incorrect.
92 Medium
92 of 118
Application +4 | -1
Equations of the asymptotes of the hyperbola 3x2+10xy+8y2+14x+22y+7=03 x^2+10 x y+8 y^2+14 x+22 y+7=0 is
View answer and explanation
Correct option: D
3x2+10xy+8y2+14x+22y+15=03 x^2+10 x y+8 y^2+14 x+22 y+15=0
Option D (3x2+10xy+8y2+14x+22y+15=03 x^2+10 x y+8 y^2+14 x+22 y+15=0) is correct because the equation of the asymptotes differs from the hyperbola by a constant, which is found using the condition for a pair of straight lines.
93 Medium
93 of 118
Application +4 | -1
The mid point of the chord 4x−3y=54x - 3y = 5 of the hyperbola 2x2−3y2=122{x^2} - 3{y^2} = 12 is
View answer and explanation
Correct option: B
(2,1)\left( {2,1} \right)
Option B (2,1) is correct because it satisfies the condition T=S1, where T is the equation of the chord with the given midpoint and S1 is the hyperbola equation with the midpoint coordinates.
94 Medium
94 of 118
Analyzing +4 | -1
For different values of α\alpha, the locus of the point of intersection of the two straight lines 3x−y−43α=0   and  3αx+αy−43=0  \sqrt 3 x - y - 4\sqrt 3 \alpha = 0\;\,\mathrm{and\;}\sqrt 3 \alpha x + \alpha y - 4\sqrt 3 = 0\; is
View answer and explanation
Correct option: A
a hyperbola with eccentricity 2
Option A (a hyperbola with eccentricity 2) is correct because the locus of the intersection point is a hyperbola, and its eccentricity is calculated to be 2.
95 Medium
95 of 118
Application +4 | -1
PP is a point on the hyperbola x281−y249=1\frac{x^2}{81}-\frac{y^2}{49}=1. The tangent at PP meets the transverse axis at T,NT, N is the foot of the perpendicular from PP to the transverse axis. If OO is the origin, then ON.OTON.OT is equal to.
View answer and explanation
Correct option: A
81
Option A (81) is correct because the product of ON and OT is calculated using the parametric coordinates of a point on the hyperbola and the equation of the tangent.
96 Medium
96 of 118
Application +4 | -1
Any point on the hyperbola (x+1)216−(y−2)24=1\frac{{{{\left( {x + 1} \right)}^2}}}{{16}} - \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1 is of the form
View answer and explanation
Correct option: D
(4  sec  θ−1,  2tan⁡θ+2)(4\mathrm{\;sec\;\theta } - 1,\mathrm{\;}2\tan \mathrm{\theta } + 2)
Option D ((4  sec;θ−1,  2tan⁡θ+2)(4\mathrm{\;sec\\;\theta } - 1,\mathrm{\;}2\tan \mathrm{\theta } + 2)) is correct because it represents a general point on the given hyperbola.
97 Medium
97 of 118
Application +4 | -1
The product of the perpendiculars from the foci on any tangent to the hyperbola x264−y29=1\frac{x^2}{64}-\frac{y^2}{9}=1 is
View answer and explanation
Correct option: B
9
Option B (9) is correct because the product of the perpendiculars from the foci to any tangent of the hyperbola is equal to b2b^2.
98 Medium
98 of 118
Analyzing +4 | -1
The locus of the point of intersection of the lines xcot⁡θ+y  cosecθ=2x\cot \mathrm{\theta } + y\;\mathrm{cosec \theta } = 2 and x  cosecθ+ycot⁡θ=6x\;\mathrm{cosec \theta } + y\cot \mathrm{\theta } = 6 is
View answer and explanation
Correct option: C
A hyperbola
Option C (A hyperbola) is correct because eliminating theta from the given equations results in the equation of a hyperbola.
99 Medium
99 of 118
Application +4 | -1
The normal at P(x1,y1)P\left(x_1, y_1\right) on the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 meets the coordinate axes at AA and BB. If OO, is the origin and ee, the eccentricity of the hyperbola, then
View answer and explanation
Correct option: A
OA=e2x1O A=e^2 x_1
Option A (OA=e2x1OA=e^2 x_1) is correct because the x-intercept of the normal is e2x1e^2x_1, which represents the distance OA.
100 Medium
100 of 118
Application +4 | -1
Any point on the hyperbola (x+1)216−(y−2)24=1\frac{{{{\left( {x + 1} \right)}^2}}}{{16}} - \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1 is of the form
View answer and explanation
Correct option: B
(4sec⁡θ−1,2tan⁡θ+2)(4\sec \mathrm{\theta } - 1,2\tan \mathrm{\theta } + 2)
Option B ((4sec⁡θ−1,2tan⁡θ+2)(4\sec \mathrm{\theta } - 1,2\tan \mathrm{\theta } + 2)) is correct because it represents a general point on the given hyperbola after considering the shift in the center.
101 Medium
101 of 118
Application +4 | -1
The condition that the straight line lx+my=nl x+m y=n may be a normal to the hyperbola b2x2−a2y2=a2b2b^2 x^2-a^2 y^2=a^2 b^2 is given by
View answer and explanation
Correct option: A
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 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 derived by comparing the general normal equation to the hyperbola with the given line equation.
102 Medium
102 of 118
Application +4 | -1
Length of the straight line x−3y=1x - 3y = 1 intercepted by the hyperbola x2−4y2=1{x^2} - 4{y^2} = 1 is
View answer and explanation
Correct option: B
6510\frac{6}{5}\sqrt {10}
Option B (6510\frac{6}{5}\sqrt {10}) is correct because it is the distance between the intersection points of the line and the hyperbola.
103 Medium
103 of 118
Application +4 | -1
The value of mm, for which the line y=mx+2533y=m x+\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: B
−2/3-2 / \sqrt{3}
Option D (2x+3y=252 x+\sqrt{3} y=25) is correct because substituting the point (8, 3√3) into the equation satisfies it, and it is the normal to the hyperbola at that point.
104 Medium
104 of 118
Application +4 | -1
The area of the triangle formed by any tangent to the hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 with its asymptotes is
View answer and explanation
Correct option: D
abab
Option D (abab) is correct because the area of the triangle formed by the tangent to the hyperbola and its asymptotes is constant and equal to abab.
105 Medium
105 of 118
Application +4 | -1
A common tangent to 9x2−16y2=1449{x^2} - 16{y^2} = 144 and x2+y2=9{x^2} + {y^2} = 9, is
View answer and explanation
Correct option: B
y=327x+157y = 3\sqrt {\frac{2}{7}} x + \frac{{15}}{{\sqrt 7 }}
Option B (y=327x+157y = 3\sqrt {\frac{2}{7}} x + \frac{{15}}{{\sqrt 7 }}) is correct because it represents a line that is tangent to both the hyperbola and the circle.
106 Medium
106 of 118
Analyzing +4 | -1
Two conics x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 and x2=−abyx^2=-\frac{a}{b} y intersect, if
View answer and explanation
Correct option: B
0<a<120 < a < \frac{1}{2}
Option A (0<a<120 < a < \frac{1}{2}) is correct because the condition for intersection requires the discriminant of the resulting quadratic equation to be positive, leading to the inequality 0<a<120 < a < \frac{1}{2}.
107 Medium
107 of 118
Analyzing +4 | -1
The circle drawn on the line segment joining the foci of the hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 as diameter cuts the asymptotes at
View answer and explanation
Correct option: D
(±a,±  b)\left( { \pm a, \pm \;b} \right)
Option C ((±  b,±  a)\left( { \pm \;b, \pm \;a} \right)) is correct because the circle intersects the asymptotes at these points.
108 Medium
108 of 118
Application +4 | -1
The locus of point of intersection of straight lines xa+yb=λ\frac{x}{a}+\frac{y}{b}=\lambda and xa−yb=1λ\frac{x}{a}-\frac{y}{b}=\frac{1}{\lambda} where λ\lambda is variable, is a
View answer and explanation
Correct option: C
Hyperbola
The correct answer is Hyperbola because multiplying the two given equations eliminates λ\lambda and results in the standard form of a hyperbola.
109 Medium
109 of 118
Application +4 | -1
Find the equation of the hyperbola whose foci are (6,4)(6,4) and (−4,4)(-4,4) and eccentricity is 2 .
View answer and explanation
Correct option: A
12x2−4y2−24x+32y−127=012 x^2-4 y^2-24 x+32 y-127=0
Option A (12x2−4y2−24x+32y−127=012 x^2-4 y^2-24 x+32 y-127=0) is correct because it represents the equation of the hyperbola derived from the given foci and eccentricity.
110 Medium
110 of 118
Application +4 | -1
The centre of the hyperbola 9x2−16y2+18x+32y−151=09 x^2-16 y^2+18 x+32 y-151=0 is
View answer and explanation
Correct option: B
(−1,1)(-1,1)
Option B ((-1,1)) is correct because it is the center of the hyperbola found by completing the square and rewriting the equation in standard form.
111 Medium
111 of 118
Application +4 | -1
The product of perpendiculars drawn from any point of a hyperbola to its asymptotes is
View answer and explanation
Correct option: A
a2b2a2+b2\frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}
Option A (a2b2a2+b2\frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}) is correct because the product of the perpendiculars from any point on a hyperbola to its asymptotes is a constant value, which is given by the formula a2b2a2+b2\frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}.
112 Medium
112 of 118
Application +4 | -1
Locus of the mid-point of the chords of the hyperbola x2−y2=a2x^2-y^2=a^2, that touch the parabola y2=4axy^2=4 a x, is
View answer and explanation
Correct option: B
y2(x−a)=x3y^2(x-a)=x^3
Option B (y2(x−a)=x3y^2(x-a)=x^3) is correct because it represents the locus of the mid-point of the chords of the hyperbola that touch the parabola.
113 Medium
113 of 118
Application +4 | -1
The equation of the tangents to the conic 3x2−y2=33 x^2-y^2=3 perpendicular to the line x+3y=2x+3 y=2 is
View answer and explanation
Correct option: A
y=3x±6y=3 x \pm \sqrt{6}
Option A (y=3x±6y=3 x \pm \sqrt{6}) is correct because it represents the tangent lines with a slope perpendicular to the given line, derived from the hyperbola equation.
114 Medium
114 of 118
Analyzing +4 | -1
The curve with parametric equation x=et+e−t  y=et−  e−tx = {e^t} + {e^{ - t}}\;y = {e^t} - \;{e^{ - t}} and is
View answer and explanation
Correct option: C
A hyperbola
Option C is correct because eliminating the parameter 't' results in the equation of a hyperbola.
115 Medium
115 of 118
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 assumptions, then the equation of the transverse axis of the hyperbola is
View answer and explanation
Correct option: B
x−7y−5=0x-7 y-5=0
Option B (x−7y−5=0x-7y-5=0) is correct because the transverse axis is the bisector of the angle between the asymptotes that contains the origin.
116 Medium
116 of 118
Analyzing +4 | -1
If a variable straight line xcos⁡α+ysin⁡α=px \cos \alpha+y \sin \alpha=p which is a chord of hyperbola x2a2−y2b2=1(b>a)\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(b>a) subtends a right angle at the centre of the hyperbola, then it always touches a fixed circle whose radius, is
View answer and explanation
Correct option: C
abb2−a2\frac{a b}{\sqrt{b^2-a^2}}
Option C (abb2−a2\frac{a b}{\sqrt{b^2-a^2}}) is correct because it represents the radius of the circle that the variable straight line always touches when it subtends a right angle at the center of the hyperbola.
117 Medium
117 of 118
Application +4 | -1
The equation of the chord of the hyperbola 25x2−16y2=40025 x^2-16 y^2=400 that is bisected at point (5,3)(5,3) is:
View answer and explanation
Correct option: B
125x−48y=481125 x-48 y=481
Option B (125x−48y=481125 x-48 y=481) is correct because it is the equation of the chord of the hyperbola bisected at the point (5,3), found using the formula T=S1.
118 Medium
118 of 118
Application +4 | -1
If the tangent at the point P(asec⁡θ,btan⁡θ)P(a \sec \theta, b \tan \theta) to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 passes through the point where a directrix of the hyperbola meets the positive side of the transverse axis, then θ\theta is equal to
Where ee is the eccentricity of the hyperbola.
View answer and explanation
Correct option: A
cos⁡−1(1/e)\cos ^{-1}(1 / e)
Option A (cos⁡−1(1/e)\cos^{-1}(1/e)) is correct because substituting the directrix point into the tangent equation and solving for θ\theta gives cos⁡θ=1/e\cos \theta = 1/e.
Revision notes and key takeaways

Review HYPERBOLA in your textbook or class notes, then use this set to check your understanding.

Showing 91-118 of 118 questions

Your practice

0 / 118 answered · 118 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.