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JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Mixed difficulty

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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 301-330 of 429 questions

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301 Easy
301 of 429
Application +4 | -1
If x2a+7+y25−a=1\frac{x^2}{a+7}+\frac{y^2}{5-a}=1 represents a hyperbola, then
View answer and explanation
Correct option: D
a<−7a < -7 or a>5a > 5
Option D (a<−7a < -7 or a>5a > 5) is correct because for the equation to represent a hyperbola, either a+7>0a+7 > 0 and 5−a<05-a < 0 or a+7<0a+7 < 0 and 5−a>05-a > 0 must be true.
302 Easy
302 of 429
Application +4 | -1
The eccentricity of the hyperbola 19993(x2−y2)=1\frac{\sqrt{1999}}{3}\left(x^2-y^2\right)=1 is
View answer and explanation
Correct option: B
2\sqrt{2}
Option B (2\sqrt{2}) is correct because the given hyperbola is a rectangular hyperbola (since a=b), and the eccentricity of any rectangular hyperbola is always 2\sqrt{2}.
303 Easy
303 of 429
Application +4 | -1
If the normal at PP to the rectangular hyperbola x2−y2=4x^2-y^2=4 meets the axes of xx and yy in GG and gg respectively and CC is the centre of the hyperbola, then 2PC=2 P C=
View answer and explanation
Correct option: C
GgG g
Option C (GgG g) is correct because the distance between the points where the normal intersects the x and y axes is twice the distance from the center of the hyperbola to point P.
304 Easy
304 of 429
Application +4 | -1
If the origin is shifted to the point (3,72)\left(3, \frac{7}{2}\right) and the axes are rotated through an angle θ\theta in a clockwise direction, so that the equation of the given hyperbola changes to the standard form x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, then θ\theta is
View answer and explanation
Correct option: B
tan⁡−1(34)\tan ^{-1}\left(\frac{3}{4}\right)
Option B (tan⁡−1(34)\tan^{-1}(\frac{3}{4})) is correct because the angle of rotation is the angle whose tangent is the slope of the transverse axis.
305 Easy
305 of 429
Application +4 | -1
If α+β=3π\alpha+\beta=3 \pi, then the chord joining the points α\alpha and β\beta for the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 passes through
View answer and explanation
Correct option: B
centre
Option B (centre) is correct because when α+β=3π\alpha + \beta = 3\pi, the equation of the chord simplifies to one that always passes through the origin (0,0), which is the center of the hyperbola.
306 Medium
306 of 429
Application +4 | -1
If the chord of contact of tangents from a point PP to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 subtends a right angle at the centre, then the locus of P{P} is
View answer and explanation
Correct option: B
An ellipse
Option B is correct because the locus of P is an ellipse, derived from the condition that the chord of contact subtends a right angle at the center.
307 Easy
307 of 429
Application +4 | -1
The hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 does not have mutually ⊥r\perp r tangents, then possible values of eccentricities are
a) 32\frac{3}{2}\qquad&nbsp; b) 87\frac{8}{7}\qquad&nbsp;c) 53\frac{5}{3}\qquad&nbsp;d) 75\frac{7}{5}
View answer and explanation
Correct option: C
Only a,c are correct
Option C (Only a,c are correct) is correct because the condition for the hyperbola not to have mutually perpendicular tangents is that e > 2\sqrt{2}, and only 32\frac{3}{2} and 53\frac{5}{3} satisfy this.
308 Medium
308 of 429
Application +4 | -1
Find the equation of the hyperbola whose directrix is 2x+y=12 x+y=1, focus is (1,1)(1,1) and eccentricity is 3\sqrt{3}.
View answer and explanation
Correct option: A
7x2+12xy−2y2−2x+4y−7=07 x^2+12 x y-2 y^2-2 x+4 y-7=0
Option A (7x2+12xy−2y2−2x+4y−7=07 x^2+12 x y-2 y^2-2 x+4 y-7=0) is correct because it is the equation of the hyperbola derived using the focus-directrix definition and the given eccentricity.
309 Easy
309 of 429
Application +4 | -1
The eccentricity of the hyperbola 4x2−9y2=164 x^2-9 y^2=16 is
View answer and explanation
Correct option: C
13/3\sqrt{13} / 3
Option C (13/3\sqrt{13} / 3) is correct because it is the calculated eccentricity of the given hyperbola after converting it to standard form.
310 Medium
310 of 429
Application +4 | -1
If the eccentricity of the hyperbola x2−y2sec⁡2α=5x^2-y^2 \sec ^2 \alpha=5 is 3\sqrt{3} times the eccentricity of the ellipse x2sec⁡2α+y2=25x^2 \sec ^2 \alpha+y^2=25, then a value of α\alpha is
View answer and explanation
Correct option: B
π4\frac{\pi}{4}
Option B (π4\frac{\pi}{4}) is correct because it is the value of α\alpha that satisfies the given condition relating the eccentricities of the hyperbola and the ellipse.
311 Easy
311 of 429
Analyzing +4 | -1
Consider a branch of the hyperbola x2−2y2−22x−42y−6=0x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0 with vertex at the point AA. Let BB be one of the end points of its latusrectum. If C\mathrm{C} is the focus of the hyperbola nearest to the point AA, then the area of the triangle ABCA B C, is
View answer and explanation
Correct option: B
32−1\sqrt{\frac{3}{2}}-1
Option B (32−1\sqrt{\frac{3}{2}}-1) is correct because it is the area of the triangle ABCABC calculated using the coordinates of the vertex, focus, and endpoint of the latus rectum.
312 Medium
312 of 429
Application +4 | -1
The locus of a point P(α,β)P\left( {\mathrm{\alpha },\mathrm{\beta }} \right) moving under the condition that the line y=αx+βy = \mathrm{\alpha }x + \mathrm{\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: A
A hyperbola
Option A (A hyperbola) is correct because the locus of the point is derived from the condition of tangency to a hyperbola, resulting in another hyperbola.
313 Medium
313 of 429
Analyzing +4 | -1
The condition that a straight line with slope mm will be normal to parabola y2=4axy^2=4 a x as well as a tangent to rectangular hyperbola x2−y2=a2x^2-y^2=a^2 is
View answer and explanation
Correct option: D
m6+4m4+3m2+1=0m^6+4 m^4+3 m^2+1=0
Option D (m6+4m4+3m2+1=0m^6+4 m^4+3 m^2+1=0) is correct because it represents the condition derived by equating the normal to the parabola and the tangent to the rectangular hyperbola.
314 Easy
314 of 429
Application +4 | -1
The equation of the normal at the point (asec⁡θ,  btan⁡θ)  (a\sec \mathrm{\theta },\;b\tan \mathrm{\theta })\; of the curve   b2x2−a2y2=a2b2  \;{b^2}{x^2} - {a^2}{y^2} = {a^2}{b^2}\; is
View answer and explanation
Correct option: C
axsec⁡θ+bytan⁡θ=a2+b2\frac{{ax}}{{\sec \mathrm{\theta }}} + \frac{{by}}{{\tan \mathrm{\theta }}} = {a^2} + {b^2}
The correct answer is axsec⁡θ+bytan⁡θ=a2+b2\frac{{ax}}{{\sec \mathrm{\theta }}} + \frac{{by}}{{\tan \mathrm{\theta }}} = {a^2} + {b^2} because this is the standard form of the normal equation to a hyperbola at the given point.
315 Medium
315 of 429
Application +4 | -1
Tangent to hyperbola xy=c2x y=c^2 at point PP intersects the xx-axis at TT and the yy-axis at T′T^{\prime}. Normal to hyperbola at P{P} intersects the x{x}-axis at N{N} and the y{y}-axis at N′{N}^{\prime}. If the area of the triangles PNTPNT and PN′T′PN^{\prime}T^{\prime} are Δ\Delta and Δ′\Delta^{\prime} respectively then 1Δ+1Δ′\frac{1}{\Delta}+\frac{1}{\Delta^{\prime}}, is equal to
View answer and explanation
Correct option: B
2c2\frac{2}{{c}^2}
Option B (2c2\frac{{2}}{{c^2}}) is correct because, after finding the coordinates of T, T', N, and N', and calculating the areas of the triangles PNT and PN'T', the expression 1Δ+1Δ′\frac{{1}}{{\Delta}} + \frac{{1}}{{\Delta'}} simplifies to 2c2\frac{{2}}{{c^2}}.
316 Easy
316 of 429
Understanding +4 | -1
The eccentricity of the curve x2−y2=1x^2-y^2=1 is
View answer and explanation
Correct option: D
2\sqrt{2}
Option D (2\sqrt{2}) is correct because the eccentricity of a rectangular hyperbola (where a=b) is always 2\sqrt{2}.
317 Easy
317 of 429
Application +4 | -1
The equation x2+4xy+y2+2x+4y+2=0x^2+4 x y+y^2+2 x+4 y+2=0 represents
View answer and explanation
Correct option: C
A hyperbola
Option C (A hyperbola) is correct because the discriminant (h2−abh^2 - ab) is positive, indicating a hyperbola.
318 Easy
318 of 429
Application +4 | -1
The eccentricity of the conic defined by ∣(x−1)2+(y−2)2−(x−5)2+(y−5)2∣=3\left|\sqrt{(x-1)^2+(y-2)^2}-\sqrt{(x-5)^2+(y-5)^2}\right|=3 is K\mathrm{K}, then 3 K=3 \mathrm{~K}=
View answer and explanation
Correct option: D
5
Option D (5) is correct because the eccentricity of the hyperbola is calculated using the given foci and transverse axis length, resulting in K = 5/3, and 3K = 5.
319 Easy
319 of 429
Application +4 | -1
If the slope of a tangent to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 is 222 \sqrt{2}, then the eccentricity ee of the hyperbola lies in the interval.
View answer and explanation
Correct option: C
(1,3](1,3]
Option C ((1,3](1,3]) is correct because the condition for the tangent to exist implies ee lies in the interval (1,3](1, 3].
320 Easy
320 of 429
Analyzing +4 | -1
From a point on the line y=x+c,cy=x+c, c (parameter), tangents are drawn to the hyperbola x22−y21=1\frac{x^2}{2}-\frac{y^2}{1}=1 such that chords of contact pass through a fixed point (x1,y1)\left(x_1, y_1\right). Then, x1y1\frac{x_1}{y_1} is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because the chord of contact passes through a fixed point, and the ratio x1y1\frac{x_1}{y_1} can be determined from the equation of the chord of contact.
321 Easy
321 of 429
Application +4 | -1
The eccentricity of the hyperbola x216−y225=1\frac{x^2}{16}-\frac{y^2}{25}=1 is
View answer and explanation
Correct option: C
41/4\sqrt{41} / 4
Option C (41/4\sqrt{41} / 4) is correct because the eccentricity of the hyperbola is calculated using the formula e=1+(b2/a2)e = \sqrt{{1 + (b^2 / a^2)}}.
322 Medium
322 of 429
Analyzing +4 | -1
If the circle x2+y2=a2{x^2} + {y^2} = {a^2} intersects the hyperbola xy=c2xy = {c^2} in four points P(x1,y1),Q(x2,y2),R(x3,y3)P\left( {{x_1},{y_1}} \right),Q\left( {{x_2},{y_2}} \right),R\left( {{x_3},{y_3}} \right) and S(x4,y4)S\left( {{x_4},{y_4}} \right), then
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the sum of the x-coordinates and y-coordinates of the intersection points are zero, and the product of the x-coordinates equals the product of the y-coordinates, both equal to c4c^4.
323 Medium
323 of 429
Application +4 | -1
A common tangent to x2−2y2=18x^2-2 y^2=18 and x2+y2=9x^2+y^2=9 is
View answer and explanation
Correct option: B
y=2x+33y=\sqrt{2} x+3 \sqrt{3}
Option B (y=2x+33y=\sqrt{2} x+3 \sqrt{3}) is correct because it satisfies the tangency conditions for both the hyperbola and the circle.
324 Easy
324 of 429
Application +4 | -1
The latusractum of the hyperbola 9x2−16y2+72x−32−16=09{x^2} - 16{y^2} + 72x - 32 - 16 = 0 is
View answer and explanation
Correct option: A
92\frac{9}{2}
Option A (9/2) is correct because it is the result of calculating the latus rectum using the formula 2b2/a after rewriting the hyperbola equation in standard form.
325 Hard
325 of 429
Application +4 | -1
If the angle between the asymptotes of the hyperbola (x2a2−y2b2)\left(\frac{x^2}{a^2}-\frac{y^2}{b^2}\right) is 2θ2 \theta and ee is its eccentricity then sin⁡θ\sin \theta is equal to
View answer and explanation
Correct option: D
1−1e2\sqrt{1-\frac{1}{e^2}}
Option D (1−1e2\sqrt{1-\frac{1}{e^2}}) is correct because it correctly relates the angle between asymptotes and the eccentricity of the hyperbola.
326 Hard
326 of 429
Analyzing +4 | -1
Let P(asec⁡θ,btan⁡θ)P(a\sec \mathrm{\theta },b\tan \mathrm{\theta }) and Q(asec⁡ϕ,btan⁡ϕ),Q(a\sec \phi ,b\tan \phi ), where θ+ϕ=π2\mathrm{\theta } + \phi = \frac{\mathrm{\pi }}{2} be two points on the hyperbola x2a2−y2b2=1.\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1.. If (h,k)\left( {h,k} \right) is the point of intersection of normals at PP and QQ, then kk is equal to
View answer and explanation
Correct option: D
−[a2+b2b]- \left[ {\frac{{{a^2} + {b^2}}}{b}} \right]
Option D ( −[a2+b2b]\ - \left[ {\frac{{{{a^2} + {b^2}}}}{b}} \right]) is correct because it is the y-coordinate of the intersection point of the normals at P and Q, derived using the given condition θ + φ = π/2.
327 Easy
327 of 429
Application +4 | -1
PP is any point on the hyperbola x2−y2=a2x^2-y^2=a^2. If F1F_1 and F2F_2 are the foci of the hyperbola and PF1⋅PF2=λ(OP)2P F_1 \cdot P F_2=\lambda(O P)^2, where OO is the origin, then λ\lambda is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because by substituting the given hyperbola equation and simplifying, we find that λ\lambda equals 1.
328 Easy
328 of 429
Application +4 | -1
If a hyperbola passes through the point P(10, 16)P(10,\,16) and it has vertices at (±6, 0),( \pm 6,\,0), then the equation of the normal to it at PP is
View answer and explanation
Correct option: C
2x+5y=1002x + 5y = 100
Option B (222\sqrt 2) is correct because it is the length of the perpendicular from the origin to the normal of the curve at the given point.
329 Easy
329 of 429
Application +4 | -1
The equation of the tangent to the hyperbola 16x2−9y2=14416 x^2-9 y^2=144 at (5,8/3)(5,8 / 3), is
View answer and explanation
Correct option: B
10x−3y=1810 x-3 y=18
Option B (10x−3y=1810 x-3 y=18) is correct because it is the equation of the tangent to the hyperbola at the given point, found by substituting the point into the hyperbola's equation.
330 Easy
330 of 429
Application +4 | -1
If the normal at the point 't1t_1' to the rectangular hyperbola xy=c2x y=c^2 meets it again at the point 't2t_2', then
View answer and explanation
Correct option: A
t2=−1t13t_2=-\frac{1}{t_1{ }^3}
Option A (t2=−1t13t_2=-\frac{1}{t_1{ }^3}) is correct because it represents the relationship between the parameters of the points where the normal to a rectangular hyperbola intersects the curve again.
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