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JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Easy

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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.

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Showing 241-270 of 298 questions
241 Easy
241 of 298
Application +4 | -1
If circle x2+y2+2gx+2fy+k=0{x^2} + {y^2} + 2\mathrm{g}x + 2fy + k = 0 intersect hyperbola xy=c2xy = {c^2} at four points (xi,y1),i=1,2,3,4\left( {{x_i},{y_1}} \right),i = 1,2,3,4 then
View answer and explanation
Correct option: B
x1+x2+x3+x4=−2g{x_1} + {x_2} + {x_3} + {x_4} = - 2\mathrm{g}
Option B (x1+x2+x3+x4=−2g{x_1} + {x_2} + {x_3} + {x_4} = - 2\mathrm{g}) is correct because it represents the sum of the roots of the quartic equation formed by the intersection of the circle and hyperbola.
242 Easy
242 of 298
Application +4 | -1
PF×PG=P F \times PG=
View answer and explanation
Correct option: A
CA2C A^2
Option A (CA2CA^2) is correct because PF x PG equals a2a^2 which is the square of the semi-major axis CA.
243 Easy
243 of 298
Analyzing +4 | -1
Let S={(x,y)∈R:y21+r−x21−r=1}S=\left\{(x, y) \in R: \frac{y^2}{1+r}-\frac{x^2}{1-r}=1\right\}, where r≠±1r \neq \pm 1. Then, SS represents
View answer and explanation
Correct option: D
An ellipse whose eccentrieity is 2r+1\sqrt{\frac{2}{r+1}}, where r>1\mathrm{r}>1.
The correct answer is D. When r > 1, the equation represents an ellipse, and its eccentricity is calculated as 2r+1\sqrt{\frac{2}{r+1}}.
244 Easy
244 of 298
Application +4 | -1
The angle between the two asymptotes of the hyperbola x216−y29=1\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1 is
View answer and explanation
Correct option: C
2tan⁡−1(34)2{\tan ^{ - 1}}\left( {\frac{3}{4}} \right)
Option C (2tan⁡−1(34)2{\tan ^{ - 1}}\left( {\frac{3}{4}} \right)) is correct because it represents the angle between the two asymptotes of the given hyperbola.
245 Easy
245 of 298
Application +4 | -1
P(10,16)P(10,16) is a point on the hyperbola having vertices at (±6,0)(\pm 6,0). The equation of the normal to the hyperbola at PP is
View answer and explanation
Correct option: A
2x+5y=1002 x+5 y=100
Option A (2x+5y=1002x+5y=100) is correct because it represents the equation of the normal to the hyperbola at the given point P(10,16).
246 Easy
246 of 298
Analyzing +4 | -1
Let AA and BB be two fixed points and PP, another point in the plane, moves in such a way that k1PA+k2PB=k3k_1 P A+k_2 P B=k_3, where k1,k2k_1, k_2 and k3k_3 are real constants. The locus of PP is
Which one in the options is not true?
View answer and explanation
Correct option: D
A hyperbola if k2=−1k_2=-1 and k1,k3>0k_1, k_3>0
Option D (A hyperbola if k2=−1k_2=-1 and k1,k3>0k_1, k_3>0) is not always true. The locus is not necessarily a hyperbola under these conditions.
247 Easy
247 of 298
Application +4 | -1
The equations of the asymptotes of the hyperbola
2x2+5xy+2y2−11x−7y−4=02{x^2} + 5xy + 2{y^2} - 11x - 7y - 4 = 0 are
View answer and explanation
Correct option: C
2x2+5xy+2y2−11x−7y+5=02{x^2} + 5xy + 2{y^2} - 11x - 7y + 5 = 0
Option D (None of the above) is incorrect. Option A (2x2+5xy+2y2−11x−7y+5=02{x^2} + 5xy + 2{y^2} - 11x - 7y + 5 = 0) is correct because the pair of asymptotes and the second-degree curve differ by a constant, and this constant can be found by setting the determinant to zero.
248 Easy
248 of 298
Application +4 | -1
The equation of the hyperbola which is referred to its axes as axes of coordinate, and whose distance between the foci is 16 and eccentricity is 2\sqrt{2}, is
View answer and explanation
Correct option: B
x2−y2=32x^2-y^2=32
Option B (x2−y2=32x^2-y^2=32) is correct because given the distance between foci and eccentricity, we can determine 'a' and 'b' and construct the hyperbola's equation.
249 Easy
249 of 298
Understanding +4 | -1
The angle between the asymptotes of x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 is equal to
View answer and explanation
Correct option: A
2tan⁡−1(ba)2{\tan ^{ - 1}}\left( {\frac{b}{a}} \right)
Option A (2tan⁡−1(ba)2{\tan ^{ - 1}}\left( {\frac{b}{a}} \right)) is correct because the angle between the asymptotes of a hyperbola is given by this formula.
250 Easy
250 of 298
Analyzing +4 | -1
The locus of the point of intersection of the 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 } ( λ\lambda is a variable) is
View answer and explanation
Correct option: D
A hyperbola
Option D (A hyperbola) is correct because eliminating the variable λ results in the equation of a hyperbola.
251 Easy
251 of 298
Analyzing +4 | -1
A series of hyperbola is drawn having a common transverse axis of length 2a2 a. Then the locus of a point PP on each hyperbola, such that its distance from the transverse axis is equal to its distance from an asymptote, is
View answer and explanation
Correct option: A
(x2−y2)2=4x2(x2−a2)\left(x^2-y^2\right)^2=4 x^2\left(x^2-a^2\right)
Option A ((x2−y2)2=4x2(x2−a2)\left(x^2-y^2\right)^2=4 x^2\left(x^2-a^2\right)) is correct because it represents the locus of point PP derived from the given condition that its distance from the transverse axis equals its distance from an asymptote.
252 Easy
252 of 298
Application +4 | -1
The equation of the hyperbola referred to its axes as coordinate axes, whose latus rectum is 4 and eccentricity is 3 is
View answer and explanation
Correct option: C
16x2−2y2=116 x^2-2 y^2=1
Option C (16x2−2y2=116 x^2-2 y^2=1) is correct because given the latus rectum and eccentricity, the equation of the hyperbola can be derived as 16x2−2y2=116 x^2-2 y^2=1.
253 Easy
253 of 298
Application +4 | -1
If the line y=2x+λy = 2x + \lambda be a tangent to the hyperbola 36x2−25y2=3600,36{x^2} - 25{y^2} = 3600, then λ\lambda is equal to
View answer and explanation
Correct option: C
±16\pm 16
Option C (±16\pm 16) is correct because these are the values of λ\lambda for which the line is tangent to the hyperbola, found using the tangency condition.
254 Easy
254 of 298
Analyzing +4 | -1
If the tangents drawn from a point on the hyperbola x2−y2=a2−b2x^2-y^2=a^2-b^2 to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 make angles α\alpha and β\beta with the transverse axis of the hyperbola, then
View answer and explanation
Correct option: C
tan⁡αtan⁡β=1\tan \alpha \tan \beta=1
Option C (tan⁡αtan⁡β=1\tan \alpha \tan \beta=1) is correct because the product of the slopes of the tangents from a point on the hyperbola to the ellipse is 1.
255 Easy
255 of 298
Application +4 | -1
Let x2a2+y2b2=1,a>b\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a > b be an ellipse, whose eccentricity is 12\frac{1}{\sqrt{2}} and the length of the latusrectum is 14\sqrt{14}. Then the square of the eccentricity of x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 is:
View answer and explanation
Correct option: A
32\frac{3}{2}
Option A (32\frac{3}{2}) is correct because it is the square of the eccentricity of the hyperbola calculated using the given information about the ellipse.
256 Easy
256 of 298
Application +4 | -1
The eccentricity of the conic represented by x2−y2−4x+4y+16=0x^2-y^2-4 x+4 y+16=0, is
View answer and explanation
Correct option: B
2\sqrt{2}
Option B (2\sqrt{2}) is correct because the given equation represents a rectangular hyperbola, which always has an eccentricity of 2\sqrt{2}.
257 Easy
257 of 298
Application +4 | -1
The locus of the poles of the chords of the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 which subtend a right angle at its centre is
View answer and explanation
Correct option: C
x2a4+y2b4=1a2−1b2\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2}
Option C (x2a4+y2b4=1a2−1b2\frac{{x^2}}{{a^4}}+\frac{{y^2}}{{b^4}}=\frac{{1}}{{a^2}}-\frac{{1}}{{b^2}}) is correct because the locus of the poles can be found by homogenizing the hyperbola equation using the polar equation and then applying the condition for perpendicular lines at the center.
258 Easy
258 of 298
Analyzing +4 | -1
Eccentricity of the hyperbola 2x2−2y2−3xy+9x+12y=02 x^2-2 y^2-3 x y+9 x+12 y=0 is equal to
View answer and explanation
Correct option: A
2\sqrt{2}
The correct answer is 2\sqrt{2} because the angle between the asymptotes is 90 degrees, indicating a rectangular hyperbola, which has an eccentricity of 2\sqrt{2}.
259 Easy
259 of 298
Application +4 | -1
The equation of the normal to the hyperbola x216−y29=1\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1 at (−4,0)\left( { - 4,0} \right) is
View answer and explanation
Correct option: D
y=0y = 0
Option D (y=0y = 0) is correct because the normal to a hyperbola at (-4,0) is a vertical line, and since the point lies on the x-axis, the normal is the x-axis itself.
260 Easy
260 of 298
Analyzing +4 | -1
For the hyperbola x2cos⁡2α−y2sin⁡2α=1\frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 1, which of the following remains constant when α\alpha varies?
View answer and explanation
Correct option: D
Abscissae of foci
Option D (Abscissae of foci) is correct because the x-coordinates of the foci remain constant at ±1\pm 1 as α\alpha varies.
261 Easy
261 of 298
Application +4 | -1
What is the slope of the tangent line drawn to the hyperbola xy=a(a≠0)x y=a(a \neq 0) at the point (a,1)(a, 1) ?
View answer and explanation
Correct option: B
−1/a-1 / a
Option B (-1/a) is correct because it is the derivative of the hyperbola equation evaluated at the specified point.
262 Easy
262 of 298
Application +4 | -1
The locus of a point P(α,β)\mathrm{P}(\alpha, \beta) which is moving under the condition that the line y=αx+βy=\alpha x+\beta is a tangent 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
A hyperbola
Option B (A hyperbola) is correct because the condition for tangency to a hyperbola results in an equation that represents another hyperbola.
263 Easy
263 of 298
Application +4 | -1
The locus of the point of intersection of the tangents to the hyperbola x216−y236=1\frac{x^2}{16}-\frac{y^2}{36}=1 which are at right angles is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the locus is a circle x2+y2=16−36x^2+y^2=16-36 which is not possible, so there is no pair of tangents that are at right angles.
264 Easy
264 of 298
Application +4 | -1
The centre of the hyperbola 9x2−36x−16y2+96y−252=09 x^2-36 x-16 y^2+96 y-252=0, is
View answer and explanation
Correct option: A
(2,3)(2,3)
Option A ((2,3)(2,3)) is correct because the center is directly identified from the standard form of the hyperbola equation after completing the square.
265 Easy
265 of 298
Application +4 | -1
Acceleration of a particle moving along a straight line is a function of velocity as a=2va=2 \sqrt{v}. At t=2st=2 s, its velocity v=16 ms−1v=16 \mathrm{~ms}^{-1}. Its velocity at t=3t=3 s will be
View answer and explanation
Correct option: B
25 ms−125 \mathrm{~ms}^{-1}
Option B (25 ms−125 \mathrm{~ms}^{-1}) is correct because solving the differential equation a=2va = 2\sqrt{v} with the given initial condition yields v=(t+2)2v = (t+2)^2, and at t=3, v=25.
266 Easy
266 of 298
Application +4 | -1
The co-ordinate of focus of 9x2−16y2+18x+32y−151=09 x^2-16 y^2+18 x+32 y-151=0
View answer and explanation
Correct option: A
(−6,1)(-6,1)
Option A ((-6,1)) is correct because completing the square and transforming the equation into standard form reveals the center and allows calculation of the foci coordinates.
267 Easy
267 of 298
Application +4 | -1
If the line lx+my+n=0l x+m y+n=0 touches the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, then
View answer and explanation
Correct option: B
a2l2−b2m2=n2a^2 l^2-b^2 m^2=n^2
Option B (a2l2−b2m2=n2a^2 l^2-b^2 m^2=n^2) is correct because it represents the condition for the line to be tangent to the hyperbola.
268 Easy
268 of 298
Analyzing +4 | -1
The locus of the point of intersection of the tangents at the end-points of normal chords of the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, is
View answer and explanation
Correct option: B
a6x2−b6y2=(a2+b2)2\frac{a^6}{x^2}-\frac{b^6}{y^2}=\left(a^2+b^2\right)^2
Option B (a6x2−b6y2=(a2+b2)2\frac{a^6}{x^2}-\frac{b^6}{y^2}=\left(a^2+b^2\right)^2) is correct because it represents the locus of the point of intersection of tangents at the end-points of normal chords of the hyperbola.
269 Easy
269 of 298
Analyzing +4 | -1
Let the foci of the ellipse x216+y27=1\frac{x^2}{16}+\frac{y^2}{7}=1 and the hyperbola x2144−y2α=125\frac{x^2}{144}-\frac{y^2}{\alpha}=\frac{1}{25} coincide. Then the length of the latus rectum of the hyperbola is:
View answer and explanation
Correct option: D
2710\frac{27}{10}
Option D (2710\frac{27}{10}) is correct because the length of the latus rectum is calculated using the formula 2b2/a2b^2/a after finding the value of α\alpha by equating the foci of the ellipse and hyperbola.
270 Easy
270 of 298
Application +4 | -1
The coordinates of the centre of the hyperbola x2+3xy+2y2+2x+3y+2=0{x^2} + 3xy + 2{y^2} + 2x + 3y + 2 = 0
View answer and explanation
Correct option: A
(−1,0)(-1,0)
Option A ((-1,0)) is correct because the center of the hyperbola is found by solving the partial derivatives of the equation with respect to x and y, setting them equal to zero.
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