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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 241-270 of 429 questions

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241 Easy
241 of 429
Application +4 | -1
The hyperbola x216−y29=1\frac{x^2}{16}-\frac{y^2}{9}=1 and point P(4,3)P(4,3), then find number of tangents from PP.
View answer and explanation
Correct option: C
2
Option C (2) is correct because substituting the point P(4,3) into the hyperbola's equation results in a negative value, indicating that the point lies outside the hyperbola, thus allowing for two tangents to be drawn.
242 Easy
242 of 429
Application +4 | -1
The equation of the tangent to the conic x2−y2−8x+2y+11=0{x^2} - {y^2} - 8x + 2y + 11 = 0 at (2,1)\left( {2,1} \right) is
View answer and explanation
Correct option: D
x−2=0x - 2 = 0
Option D (x−2=0x - 2 = 0) is correct because it represents the tangent line to the conic at the given point, found by using the tangent equation formula.
243 Easy
243 of 429
Application +4 | -1
The equation of a tangent to the hyperbola 16x2−25y2−96x+100y−356=016 x^2-25 y^2-96 x+100 y-356=0, which makes an angle π/4\pi / 4 with the transverse axis, is
View answer and explanation
Correct option: A
y=x+2y=x+2
Option A (y=x+2y=x+2) is correct because it represents a tangent line to the hyperbola with a slope of 1, derived using the tangent equation y−k=m(x−h)±a2m2−b2y-k=m(x-h) \pm \sqrt{a^2 m^2-b^2}.
244 Easy
244 of 429
Application +4 | -1
What is the slope of the tangent drawn to the hyperbola xy=a,  (a≠0)xy = a,\;\left( {a \ne 0} \right) at the point (a,  1)?\left( {a,\;1} \right)?
View answer and explanation
Correct option: B
−1a- \frac{1}{a}
The correct answer is B (−1a- \frac{1}{a}) because it represents the slope of the tangent to the hyperbola at the given point, calculated using differentiation.
245 Medium
245 of 429
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.
246 Easy
246 of 429
Understanding +4 | -1
The difference of the focal distance of any point on the hyperbola 9x2−16y2=1449 x^2-16 y^2=144 is
View answer and explanation
Correct option: A
8
Option A (8) is correct because the difference of the focal distance of any point on the hyperbola is equal to 2a, where a is the semi-major axis, which is 4 in this case.
247 Easy
247 of 429
Application +4 | -1
The equation x22−λ−−y2λ−5−1=0\frac{{{x^2}}}{{2 - \lambda }}--\frac{{{y^2}}}{{\lambda - 5}} - 1 = 0, represent an ellipse, if
View answer and explanation
Correct option: B
λ<2\lambda < 2
Option B (λ<2\lambda < 2) is correct because for the equation to represent an ellipse, both denominators must be positive, leading to the condition λ<2\lambda < 2.
248 Easy
248 of 429
Application +4 | -1
If m1m_1 and m2m_2 are two values of mm for which the line y=mx+25y=m x+2 \sqrt{5} is a tangent to the hyperbola x24−y216=1\frac{x^2}{4}-\frac{y^2}{16}=1, then the value of ∣m1+1m2∣\left|m_1+\frac{1}{m_2}\right| is equal to
View answer and explanation
Correct option: A
83\frac{8}{3}
Option A (83\frac{{8}}{{3}}) is correct because it involves finding the values of m for which the line is tangent to the hyperbola and then calculating the absolute value of the expression.
249 Easy
249 of 429
Application +4 | -1
The equation of the hyperbola whose foci are (6, 4) and (−4,4)( - 4,4) and eccentricity 2, is
View answer and explanation
Correct option: D
4(x−1)225−4(y−4)275=1\frac{{4{{\left( {x - 1} \right)}^2}}}{{25}} - \frac{{4{{\left( {y - 4} \right)}^2}}}{{75}} = 1
Option D (4(x−1)225−4(y−4)275=1\frac{{4{{\left( {x - 1} \right)}^2}}}{{25}} - \frac{{4{{\left( {y - 4} \right)}^2}}}{{75}} = 1) is correct because it satisfies the given foci and eccentricity for a hyperbola centered at (1, 4).
250 Hard
250 of 429
Application +4 | -1
Tangents at any points on the hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 cut the axes at AA and BB respectively. If the rectangle OAPB,OAPB, where OO is the origin is completed, then locus of point PP is given by
View answer and explanation
Correct option: A
a2x2−b2y2=1\frac{{{a^2}}}{{{x^2}}} - \frac{{{b^2}}}{{{y^2}}} = 1
Option A (a2x2−b2y2=1\frac{{{a^2}}}{{{x^2}}} - \frac{{{b^2}}}{{{y^2}}} = 1) is correct because it represents the locus of point P, where tangents to the hyperbola cut the axes at A and B, and OAPB is a rectangle.
251 Easy
251 of 429
Application +4 | -1
If the focii of the ellipse x2k2a2+y2a2=1\frac{x^2}{k^2 a^2}+\frac{y^2}{a^2}=1 and hyperbola x2a2−y2a2=1\frac{x^2}{a^2}-\frac{y^2}{a^2}=1 coincides then k=k=?
View answer and explanation
Correct option: A
±3\pm \sqrt{3}
Option A (±3\pm \sqrt{3}) is correct because equating the focal distances of the ellipse and hyperbola leads to the equation k2a2−a2=a2+a2k^2 a^2 - a^2 = a^2 + a^2, which simplifies to k=±3k = \pm \sqrt{3}.
252 Hard
252 of 429
Analyzing +4 | -1
Let F1,F2F_1, F_2 are the foci of the hyperbola x216−y29=1\frac{x^2}{16}-\frac{y^2}{9}=1 and F3,F4F_3, F_4 are the foci of its conjugate hyperbola. If eH\mathrm{e}_{\mathrm{H}} and eC\mathrm{e}_{\mathrm{C}} are eccentricities of hyperbola and its conjugate hyperbola respectively, then the statement which holds TRUE is :
View answer and explanation
Correct option: C
Area of the quadrilateral formed by their foci is 50 sq. units.
Option C (Area of the quadrilateral formed by their foci is 50 sq. units) is correct because the area of the quadrilateral formed by the foci of the hyperbola and its conjugate is indeed 50 square units.
253 Easy
253 of 429
Application +4 | -1
The pole of the line lx+my+n=0l x+m y+n=0 with respect 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
(−a2ln,b2mn)\left(-\frac{a^2 l}{n}, \frac{b^2 m}{n}\right)
Option C ((a2ln,−b2mn)\left(\frac{{a^2 l}}{n},-\frac{{b^2 m}}{n}\right)) is correct because the pole of a line with respect to a hyperbola can be found by comparing the equation of the polar with the given line.
254 Easy
254 of 429
Application +4 | -1
The value of mm, for which y=mx+6y=m x+6 is a tangent to the hyperbola x2100−y249=1\frac{x^2}{100}-\frac{y^2}{49}=1, is
View answer and explanation
Correct option: A
17/20\sqrt{17 / 20}
Option A (17/20\sqrt{17 / 20}) is correct because it satisfies the condition for tangency of a line to a hyperbola.
255 Easy
255 of 429
Application +4 | -1
Equation of the hyperbola whose vertices are (± 3,0)\left( { \pm \,3,0} \right) and foci at (± 5,0)\left( { \pm \,5,0} \right), is
View answer and explanation
Correct option: A
16x2−9y2=14416{x^2} - 9{y^2} = 144
Option B (9x2−16y2=1449{x^2} - 16{y^2} = 144) is correct because it represents a hyperbola with vertices at (±\pm3,0) and foci at (±\pm5,0).
256 Easy
256 of 429
Application +4 | -1
A line parallel to the straight line 2x−y=02x - y = 0 is tangent to the hyperbola x24−y22=1\frac{{{x^2}}}{4} - \frac{{{y^2}}}{2} = 1 at the point (x1,y1).\left( {{x_1},{y_1}} \right). Then x12+5y12x_1^2 + 5y_1^2 is equal to
View answer and explanation
Correct option: C
6
The correct answer is C (6) because it is found by determining the tangent point on the hyperbola using the parallel line's slope and substituting it back into the hyperbola equation.
257 Medium
257 of 429
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}.
258 Easy
258 of 429
Application +4 | -1
The locus of the middle points of chords of hyperbola 3x2−2y2+4x−6y=03 x^2-2 y^2+4 x-6 y=0 parallel to y=2xy=2 x is
View answer and explanation
Correct option: A
3x−4y=43 x-4 y=4
Option A (3x−4y=43 x-4 y=4) is correct because it represents the locus of the middle points of the chords parallel to the given line, found using the T = S' method and comparing slopes.
259 Medium
259 of 429
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.
260 Easy
260 of 429
Application +4 | -1
The equation of the transverse and conjugate axis of the hyperbola 16x2−y2+64x+4y+44=016 x^2-y^2+64 x+4 y+44=0 are
View answer and explanation
Correct option: C
y=2,x+2=0y=2, x+2=0
Option C (y=2,x+2=0y=2, x+2=0) is correct because these are the equations of the transverse and conjugate axes of the given hyperbola after converting it to standard form.
261 Easy
261 of 429
Understanding +4 | -1
A general point on the hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 is
View answer and explanation
Correct option: C
(aet+e−t2,bet−e−t2)\left( {a\frac{{{e^t} + {e^{ - t}}}}{2},b\frac{{{e^t} - {e^{ - t}}}}{2}} \right) (where tt is parameter)
Option B (1) is correct because molecularity is the number of reactant molecules involved in an elementary step, and in this case, only one molecule of (CH3)3CBr(CH_3)_3CBr is involved.
262 Easy
262 of 429
Analyzing +4 | -1
If there are two points AA and BB on rectangular hyperbola xy=c2x y=c^2 such that abscissa of A=A= ordinate of BB, then the locus of point of intersection of tangents at AA and BB is
View answer and explanation
Correct option: C
y=xy=x
Option C (y=xy=x) is correct because the points A and B are symmetrical about the line y=xy=x, so the intersection of tangents at A and B will lie on y=xy=x.
263 Easy
263 of 429
Application +4 | -1
If ee and e1,{e_1}, are the eccentricities of the hyperbolas xy=c2xy = {c^2} and x2−y2=c2,{x^2} - {y^2} = {c^2},, then e2+e12{e^2} + e_1^2 is equal to
View answer and explanation
Correct option: B
4
The correct answer is 4 because both hyperbolas are rectangular hyperbolas, each having an eccentricity of 2\sqrt{2}, and the sum of the squares of their eccentricities is 4.
264 Medium
264 of 429
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.
265 Easy
265 of 429
Application +4 | -1
The length of the transverse axis of the hyperbola 3x2−4y2=323 x^2-4 y^2=32 is
View answer and explanation
Correct option: A
82/38 \sqrt{2} / \sqrt{3}
Option A (82/38 \sqrt{2} / \sqrt{3}) is correct because it represents the length of the transverse axis (2a) after rewriting the hyperbola equation in standard form.
266 Easy
266 of 429
Analyzing +4 | -1
Let the tangent drawn to the parabola y2=24xy^2=24 x at the point (α,β)(\alpha, \beta) is perpendicular to the line 2x+2y2 x+2 y =5=5. Then the normal to the hyperbola x2α2−y2β2=1\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1 at the point (α+4,β+4)(\alpha+4, \beta+4) does NOT pass through the point
View answer and explanation
Correct option: D
(15,13)(15,13)
Option D ((15,13)(15,13)) is correct because the normal to the hyperbola at the given point does not pass through this point, while it passes through the other options.
267 Easy
267 of 429
Application +4 | -1
If AA line with slope mm touches the hyperbola x225−y24=1\frac{x^2}{25}-\frac{y^2}{4}=1 and the parabola y2=20xy^2=20 x. Then the value of 25m4−4m225 m^4-4 m^2 is equal to
View answer and explanation
Correct option: C
25
Option C (25) is correct because it involves using the tangency conditions for both the hyperbola and the parabola to find the value of the expression.
268 Easy
268 of 429
Application +4 | -1
If the hyperbola x2b2−y2a2=1\frac{x^2}{b^2}-\frac{y^2}{a^2}=1 passes through the focus of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, then eccentricity of the hyperbola is
View answer and explanation
Correct option: B
3\sqrt{3}
Option B (3\sqrt{3}) is correct because by substituting the ellipse's focus into the hyperbola equation and using the ellipse's eccentricity, the hyperbola's eccentricity is found to be 3\sqrt{3}.
269 Easy
269 of 429
Application +4 | -1
If θ\theta is an angle between the two asymptotes of the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, then cos⁡(θ2)\cos \left(\frac{\theta}{2}\right) is equal to
View answer and explanation
Correct option: C
aa2+b2\frac{a}{\sqrt{a^2+b^2}}
Option C (aa2+b2\frac{a}{\sqrt{a^2+b^2}}) is correct because the angle between the asymptotes can be found using their slopes, and then cos⁡(θ2)\cos(\frac{\theta}{2}) can be expressed in terms of a and b.
270 Easy
270 of 429
Application +4 | -1
If xy=λ2−9x y=\lambda^2-9 be a rectangular hyperbola whose branches lie only in the second and fourth quadrant, then
View answer and explanation
Correct option: B
∣λ∣<3|\lambda|<3
Option B (∥λ∥<3\|\lambda\|<3) is correct because for the rectangular hyperbola to lie in the second and fourth quadrants, the product xy must be negative, implying λ2−9<0\lambda^2 - 9 < 0, which simplifies to ∥λ∥<3\|\lambda\|<3.
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