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JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Medium

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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 61-90 of 118 questions

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61 Medium
61 of 118
Application +4 | -1
The straight line x+y=2px + y = \sqrt 2 p will touch the hyperbola 4x2−9y2=36,4{x^2} - 9{y^2} = 36, if
View answer and explanation
Correct option: D
2p2=52{p^2} = 5
Option D (2p2=52{p^2} = 5) is correct because the condition for the line to touch the hyperbola leads to the equation 2p2=52{p^2} = 5.
62 Medium
62 of 118
Application +4 | -1
The locus of the middle points of the portions of the tangents of the hyperbola. x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 included between the axes is
View answer and explanation
Correct option: C
b2x2−a2y2+4x2y2=0b^2 x^2-a^2 y^2+4 x^2 y^2=0
Option B (b2x2−a2y2+4x2y2=0b^2 x^2-a^2 y^2+4 x^2 y^2=0) is correct because finding the intercepts, midpoint, and eliminating θ\theta leads to this equation.
63 Medium
63 of 118
Application +4 | -1
The locus of middle points of chords of hyperbola 3x2−2y2+4x−6y=03{x^2} - 2{y^2} + 4x - 6y = 0 parallel to y=2xy = 2x is
View answer and explanation
Correct option: A
3x−4y=43x - 4y = 4
Option A (3x−4y=43x - 4y = 4) is correct because it represents the locus of the midpoints of chords of the hyperbola parallel to the given line, derived using the T = S1 method.
64 Medium
64 of 118
Application +4 | -1
If 5x2+λy2=205{x^2} + \lambda {y^2} = 20 represents a rectangular hyperbola, then λ\lambda equals
View answer and explanation
Correct option: C
−5- 5
Option C (−5- 5) is correct because for a rectangular hyperbola, the sum of the coefficients of x² and y² must be zero.
65 Medium
65 of 118
Understanding +4 | -1
Consider the set of hyperbola xy=k,k∈R.xy = k,k \in R. Let e1{e_1} be the eccentricity when k=4k = 4 and e2{e_2} be the eccentricity when k=9k = 9, then e1−e2{e_1} - {e_2} is equal to
View answer and explanation
Correct option: B
0
The correct answer is B (0) because both hyperbolas xy=4xy=4 and xy=9xy=9 are rectangular hyperbolas, and all rectangular hyperbolas have the same eccentricity of 2\sqrt{2}.
66 Medium
66 of 118
Application +4 | -1
The shortest distance between the line y=xy=x and the hyperbola x29−y24=1\frac{x^2}{9}-\frac{y^2}{4}=1 is
View answer and explanation
Correct option: C
52\sqrt{\frac{5}{2}}
Option C (52\sqrt{\frac{5}{2}}) is correct because it represents the shortest distance calculated using the tangent to the hyperbola parallel to the line y=x.
67 Medium
67 of 118
Application +4 | -1
If the chords of contacts of the tangents from the points (x1,y1)\left(x_1, y_1\right) and (x2,y2)\left(x_2, y_2\right) to the hyperbola 2x2−3y2=62 x^2-3 y^2=6 are at right angle, then 4x1x2+9y1y24 x_1 x_2+9 y_1 y_2 is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because the chords of contact are perpendicular, leading to the condition 4x1x2+9y1y2=04x_1x_2 + 9y_1y_2 = 0.
68 Medium
68 of 118
Application +4 | -1
PP is a point on the hyperbola x2a2−y2b2=1,  N\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1,\;N is the foot of the perpendicular from PP on the transverse axis. The tangent to the hyperbola at PP meets the transverse axis at TT. If OO is the centre of the hyperbola, then OT.ONOT.ON is equal to
View answer and explanation
Correct option: B
a2{a^2}
Option B (a2{a^2}) is correct because the product of OT and ON simplifies to the square of the semi-major axis of the hyperbola.
69 Medium
69 of 118
Application +4 | -1
If tt is a parameter, then x=a(t+1t),y=b(t−1t)x = a\left( {t + \frac{1}{t}} \right),y = b\left( {t - \frac{1}{t}} \right) represents
View answer and explanation
Correct option: D
A hyperbola
Option D (A hyperbola) is correct because eliminating the parameter 't' from the given parametric equations results in the equation of a hyperbola.
70 Medium
70 of 118
Application +4 | -1
The equation of tangent to 4x2−9y2=364 x^2-9 y^2=36 which are perpendicular to straight line 5x+2y−10=05 x+2 y-10=0 are
View answer and explanation
Correct option: D
None of these
Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because the mean deviation is correctly calculated using the provided formula, and the reason accurately explains the method.
71 Medium
71 of 118
Analyzing +4 | -1
For some θ∈(0,π2)\theta \in\left(0, \frac{\pi}{2}\right) if eccentricity of the hyperbola x2−y2cosec⁡2θ=8x^2-y^2 \operatorname{cosec}^2 \theta=8 is 7\sqrt{7} times the eccentricity of the ellipse x2+y2cosec⁡2θ=4x^2+y^2 \operatorname{cosec}^2 \theta=4, then the length of the latus rectum of the ellipse is
View answer and explanation
Correct option: B
32\frac{3}{2}
Option B (32\frac{3}{2}) is correct because after finding the relationship between the eccentricities and solving for sin⁡2θ\sin^2 \theta, the length of the latus rectum of the ellipse can be calculated.
72 Medium
72 of 118
Application +4 | -1
The asymptotes of the hyperbola xy=hx+kyxy = hx + ky are
View answer and explanation
Correct option: A
x=k,y=hx = k,y = h
Option A (x=k,y=hx = k, y = h) is correct because, by rearranging the equation and shifting the origin, the asymptotes can be identified as x = k and y = h.
73 Medium
73 of 118
Analyzing +4 | -1
The equation 16x2−3y2−32x+12y−44=016 x^2-3 y^2-32 x+12 y-44=0 represents a hyperbola
View answer and explanation
Correct option: D
Whose eccentricity is 193\sqrt{\frac{19}{3}}
Option D (Whose eccentricity is 193\sqrt{\frac{19}{3}}) is correct because after converting the equation to standard form, the eccentricity is calculated using the formula e=1+(b2/a2)e = \sqrt{1 + (b^2/a^2)}.
74 Medium
74 of 118
Application +4 | -1
The equation of a hyperbola, conjugate to the hyperbola x2+3xy+2y2+2x+3y=0x^2+3 x y+2 y^2+2 x+3 y=0 is
View answer and explanation
Correct option: B
x2+3xy+2y2+2x+3y+2=0x^2+3 x y+2 y^2+2 x+3 y+2=0
Option B (x2+3xy+2y2+2x+3y+2=0x^2+3 x y+2 y^2+2 x+3 y+2=0) is correct because the conjugate hyperbola can be found by using the relationship H + C = 2A, where H is the original hyperbola, C is the conjugate hyperbola, and A is the pair of asymptotes.
75 Medium
75 of 118
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.
76 Medium
76 of 118
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.
77 Medium
77 of 118
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.
78 Medium
78 of 118
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.
79 Medium
79 of 118
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.
80 Medium
80 of 118
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}}.
81 Medium
81 of 118
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.
82 Medium
82 of 118
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.
83 Medium
83 of 118
Analyzing +4 | -1
If tangent to the hyperbola x24−y2=1\frac{x^2}{4}-y^2=1 cut the circle x2+y2=16x^2+y^2=16 at two distinct point AA and BB, then the locus of mid-point of AA and BB is
View answer and explanation
Correct option: D
x2+y2=4x2−y2x^2+y^2=\sqrt{4 x^2-y^2}
Option D (x2+y2=4x2−y2x^2+y^2=\sqrt{4 x^2-y^2}) is correct because it represents the locus of the midpoint of the chord formed by the tangent to the hyperbola intersecting the circle.
84 Medium
84 of 118
Analyzing +4 | -1
The tangent to the hyperbola xy=c2x y=c^2 at point P(t)P(t) intersects the XX-axis at TT and the YY-axis. at T′T^{\prime}. The normal to the hyperbola at P(t)P(t) intersects the XX-axis at NN and the YY-axis at N′N^{\prime}. The areas of the triangles PNTP N T and PN′T′P N^{\prime} T^{\prime} are Δ\Delta and Δ′\Delta^{\prime} respectively, then 1Δ+1Δ′\frac{1}{\Delta}+\frac{1}{\Delta^{\prime}} is
View answer and explanation
Correct option: C
depends on cc
The correct answer is that the expression depends on cc because the areas of the triangles PNTPNT and PN′T′PN'T' are functions of cc, and their reciprocals sum to 2c2\frac{2}{c^2}.
85 Medium
85 of 118
Application +4 | -1
Equation of the chord of the hyperbola 25x2−16y2=40025 x^2-16 y^2=400 which is bisected at the point (6,2)(6,2) is
View answer and explanation
Correct option: B
75x−16y=41875 x-16 y=418
Option B (75x−16y=41875 x-16 y=418) is correct because it is found by applying the equation of the chord bisected at a given point (T = S1).
86 Medium
86 of 118
Analyzing +4 | -1
If the chords of contact of tangents drawn from PP to the hyperbola x2−y2=a2{x^2} - {y^2} = {a^2} and its auxiliary circle are at right angle, then PP lies on
View answer and explanation
Correct option: C
x2−y2=0{x^2} - {y^2} = 0
Option C (x2−y2=0{x^2} - {y^2} = 0) is correct because the chords of contact being at right angles implies that the point P lies on the asymptotes of the hyperbola.
87 Medium
87 of 118
Application +4 | -1
The product of perpendicular distances from any point on the hyperbola 9x2−16y2=1449{x^2} - 16{y^2} = 144 to its asymptotes is
View answer and explanation
Correct option: B
14425\frac{{144}}{{25}}
Option B (14425\frac{{144}}{{25}}) is correct because it represents the product of the perpendicular distances from any point on the given hyperbola to its asymptotes.
88 Medium
88 of 118
Understanding +4 | -1
The equation ∣x2+(y−1)2−x2+(y+1)2∣=k\left| {\sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} - \sqrt {{x^2} + {{\left( {y + 1} \right)}^2}} } \right| = k will represent a hyperbola for
View answer and explanation
Correct option: A
k∈(0,2)k \in \left( {0,2} \right)
Option A (k∈(0,2)k \in \left( {0,2} \right)) is correct because the equation represents a hyperbola when the difference between the distances to the foci is less than the distance between the foci.
89 Medium
89 of 118
Application +4 | -1
A hyperbola has the asymptotes x+2y=3x + 2y = 3 and x−y=0x - y = 0 and passes through (2,1)\left( {2,1} \right). Its centre is
View answer and explanation
Correct option: C
(1,1)\left( {1,1} \right)
Option C ((1,1)) is correct because the center of the hyperbola is the intersection point of its asymptotes.
90 Medium
90 of 118
Application +4 | -1
Normal at point (5,3)(5,3) to the rectangular hyperbola xy−y−2x−2=0x y-y-2 x-2=0 meets the curve at the point whose coordinates are
View answer and explanation
Correct option: D
(34,−14)\left(\frac{3}{4},-14\right).
Option D ((34,−14)\left(\frac{{3}}{{4}}, -14\right)) is correct because it involves finding the equation of the normal and then finding the intersection point with the hyperbola.
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