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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 331-360 of 429 questions

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331 Easy
331 of 429
Analyzing +4 | -1
An ellipse intersects the hyperbola 2x2−2y2=12 x^2-2 y^2=1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then
View answer and explanation
Correct option: A
Equation of the ellipse is x2+2y2=2x^2+2 y^2=2 with foci at (±1,0)(\pm 1,0)
Option A (Equation of the ellipse is x2+2y2=2x^2+2 y^2=2 with foci at (±1,0)(\pm 1,0)) is correct because it satisfies the orthogonality condition and the eccentricity relationship with the given hyperbola.
332 Easy
332 of 429
Application +4 | -1
If x2λ+3+y22−λ=1\frac{x^2}{\lambda+3}+\frac{y^2}{2-\lambda}=1 represents a hyperbola, then
View answer and explanation
Correct option: C
λ∉(−3,2)\lambda \notin(-3,2)
Option C (λ∉(−3,2)\lambda \notin(-3,2)) is correct because for the given equation to represent a hyperbola, λ+3\lambda+3 and 2−λ2-\lambda must have opposite signs, leading to this condition.
333 Easy
333 of 429
Analyzing +4 | -1
The locus of the point of intersection of the straight lines, tx−2y−3t=0\mathrm{tx}-2 \mathrm{y}-3 \mathrm{t}=0 x−2ty+3=0(t∈R)x-2 t y+3=0(t \in R), is :
View answer and explanation
Correct option: D
a hyperbola with the length of conjugate axis 3
Option D (a hyperbola with the length of conjugate axis 3) is correct because the locus is a hyperbola, and the length of the conjugate axis is correctly calculated as 3.
334 Easy
334 of 429
Application +4 | -1
The eccentricity of the hyperbola 5x2−4y2+20x+8y=45{x^2} - 4{y^2} + 20x + 8y = 4 is
View answer and explanation
Correct option: B
32\frac{3}{2}
Option B (32\frac{3}{2}) is correct because it is the eccentricity calculated after converting the given hyperbola equation into standard form.
335 Easy
335 of 429
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.
336 Easy
336 of 429
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.
337 Easy
337 of 429
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}}.
338 Medium
338 of 429
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.
339 Medium
339 of 429
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}.
340 Medium
340 of 429
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).
341 Medium
341 of 429
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.
342 Easy
342 of 429
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.
343 Medium
343 of 429
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.
344 Easy
344 of 429
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).
345 Easy
345 of 429
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.
346 Medium
346 of 429
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.
347 Easy
347 of 429
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.
348 Easy
348 of 429
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.
349 Easy
349 of 429
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.
350 Medium
350 of 429
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.
351 Easy
351 of 429
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.
352 Medium
352 of 429
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.
353 Medium
353 of 429
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.
354 Easy
354 of 429
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.
355 Easy
355 of 429
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.
356 Medium
356 of 429
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.
357 Medium
357 of 429
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.
358 Medium
358 of 429
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.
359 Easy
359 of 429
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.
360 Easy
360 of 429
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.
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