Skip to main content
Back to HYPERBOLA

Your Progress

0 / 429 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Mixed difficulty

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 151-180 of 429 questions

Preparing practice and saved progress…

151 Easy
151 of 429
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.
152 Medium
152 of 429
Application +4 | -1
Let H:−x2a2+y2 b2=1\mathrm{H}: \frac{-\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 be the hyperbola, whose eccentricity is 3\sqrt{3} and the length of the latus rectum is 434 \sqrt{3}. Suppose the point (α,6),α>0(\alpha, 6), \alpha > 0 lies on HH. If β\beta is the product of the focal distances of the point (α,6)(\alpha, 6), then α2+β\alpha^2+\beta is equal to :
View answer and explanation
Correct option: B
171
Option B (171) is correct because by finding the equation of the hyperbola and the focal distances, we can calculate α2+β\alpha^2 + \beta to be 171.
153 Medium
153 of 429
Analyzing +4 | -1
If chords of the hyperbola x2−y2=a2{x^2} - {y^2} = {a^2} touch the parabola y2=4ax.{y^2} = 4ax. Then, the locus of the middle points of these chord is
View answer and explanation
Correct option: B
y2(x−a)=x3{y^2}\left( {x - a} \right) = {x^3}
Option B (y2(x−a)=x3{y^2}({x - a}) = {x^3}) is correct because it represents the locus of the midpoints of the chords of the hyperbola that touch the given parabola.
154 Medium
154 of 429
Application +4 | -1
What is the equation of the chord of hyperbola 25x2−16y2=40025 x^2-16 y^2=400 whose mid-point is (5,3)(5,3) ?
View answer and explanation
Correct option: B
125x−48y=481125 x-48 y=481
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 for the line to be a normal to the hyperbola, derived by comparing coefficients.
155 Easy
155 of 429
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.
156 Easy
156 of 429
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.
157 Easy
157 of 429
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.
158 Easy
158 of 429
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).
159 Medium
159 of 429
Analyzing +4 | -1
Equation 1r=18+38cos⁡θ\frac{1}{r} = \frac{1}{8} + \frac{3}{8}\cos \mathrm{\theta } represents
View answer and explanation
Correct option: B
A hyperbola
Option B (A hyperbola) is correct because the given polar equation has an eccentricity greater than 1, indicating a hyperbola.
160 Easy
160 of 429
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.
161 Easy
161 of 429
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.
162 Easy
162 of 429
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}.
163 Easy
163 of 429
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.
164 Easy
164 of 429
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.
165 Easy
165 of 429
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.
166 Easy
166 of 429
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.
167 Easy
167 of 429
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}}.
168 Medium
168 of 429
Application +4 | -1
The line lx+my+n=0l x+m y+n=0 will be a normal to the hyperbola b2x2−a2y2=a2b2b^2 x^2-a^2 y^2=a^2 b^2, if
View answer and explanation
Correct option: C
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 C (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}) is correct because it represents the condition for the line to be a normal to the hyperbola.
169 Easy
169 of 429
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.
170 Easy
170 of 429
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.
171 Easy
171 of 429
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.
172 Easy
172 of 429
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.
173 Medium
173 of 429
Application +4 | -1
Let the eccentricity of the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 be the reciprocal to that of the ellipse x2+4y2=4x^2+4 y^2=4. If the hyperbola passes through a focus of the ellipse, then the equation of the hyperbola, is
View answer and explanation
Correct option: B
x2−3y2=3x^2-3 y^2=3
Option B (x2−3y2=3x^2-3 y^2=3) is correct because the hyperbola passes through a focus of the ellipse and has the reciprocal eccentricity.
174 Easy
174 of 429
Application +4 | -1
Latusrectum of the conic satisfying the differential equation xdy+ydx=0x d y+y d x=0 and passing through the point (2,8)(2,8) is
View answer and explanation
Correct option: C
828 \sqrt{2}
Option C (828 \sqrt{2}) is correct because the differential equation leads to the equation of a rectangular hyperbola, and the length of the latus rectum for xy=c2 is 2c22c\sqrt{2}.
175 Easy
175 of 429
Application +4 | -1
The equations of the transverse and conjugate axes of a hyperbola are respectively x+2y−3=0,2x−y+4=0x+2 y-3=0,2 x-y+4=0, and their respective lengths are 2\sqrt{2} and 2/32 / \sqrt{3}. The equation of the hyperbola is
View answer and explanation
Correct option: B
25(2x−y+4)2−35(x+2y−3)2=1\frac{2}{5}(2 x-y+4)^2-\frac{3}{5}(x+2 y-3)^2=1
Option B (25(2x−y+4)2−35(x+2y−3)2=1\frac{2}{5}(2 x-y+4)^2-\frac{3}{5}(x+2 y-3)^2=1) is correct because it correctly uses the given transverse and conjugate axes equations and their lengths to form the hyperbola equation.
176 Medium
176 of 429
Application +4 | -1
The coordinates of a point common to a directrix and an asymptote of the hyperbola x225−y216=1\frac{x^2}{25}-\frac{y^2}{16}=1 are
View answer and explanation
Correct option: D
(−2541,2041)\left(\frac{-25}{\sqrt{41}}, \frac{20}{\sqrt{41}}\right)
Option D ((−2541,2041)\left(\frac{{-25}}{{\sqrt{41}}}, \frac{{20}}{{\sqrt{41}}}\right)) is correct because it is the intersection point of the directrix and asymptote of the given hyperbola.
177 Easy
177 of 429
Application +4 | -1
The points of the intersection of the curves whose parametric equations are x=t2+1,y=2tx=t^2+1, y=2 t and x=2s,y=2sx=2 s, y=\frac{2}{s} is given by
View answer and explanation
Correct option: B
(2,2)(2,2)
Option B ((2,2)) is correct because by equating the parametric equations and solving for the intersection point, we find that x=2 and y=2.
178 Medium
178 of 429
Application +4 | -1
A tangent to a hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 intercepts a length of unity from each of the coordinate axes, then the point (a,b)(a, b) lies on the rectangular hyperbola
View answer and explanation
Correct option: B
x2−y2=1x^2-y^2=1
Option B (x2−y2=1x^2-y^2=1) is correct because the tangent intercepts unity from each axis, leading to the condition that a=sec⁡θa = \sec \theta and b=tan⁡θb = \tan \theta, which satisfies the equation x2−y2=1x^2 - y^2 = 1.
179 Easy
179 of 429
Application +4 | -1
The equation of the director circle of the hyperbola x216−y24=1\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{4} = 1, is given by
View answer and explanation
Correct option: D
x2+y2=12{x^2} + {y^2} = 12
Option D (x2+y2=12{x^2} + {y^2} = 12) is correct because the equation of the director circle of a hyperbola is given by x2+y2=a2−b2{x^2} + {y^2} = {a^2} - {b^2}.
180 Easy
180 of 429
Understanding +4 | -1
The eccentricity of the curve x2−y2=a2x^2-y^2=a^2 is
View answer and explanation
Correct option: B
2\sqrt{2}
Option B (2\sqrt{2}) is correct because the eccentricity of a rectangular hyperbola (where a=b) is always 2\sqrt{2}.
Revision notes and key takeaways

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

Showing 151-180 of 429 questions

Your practice

0 / 429 answered · 429 remaining · 0% accuracy

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

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

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.