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JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Easy

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Showing 181-210 of 298 questions

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181 Easy
181 of 298
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.
182 Easy
182 of 298
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).
183 Easy
183 of 298
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.
184 Easy
184 of 298
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.
185 Easy
185 of 298
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.
186 Easy
186 of 298
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.
187 Easy
187 of 298
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.
188 Easy
188 of 298
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.
189 Easy
189 of 298
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.
190 Easy
190 of 298
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.
191 Easy
191 of 298
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.
192 Easy
192 of 298
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}.
193 Easy
193 of 298
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.
194 Easy
194 of 298
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.
195 Easy
195 of 298
Application +4 | -1
The equations of the tangents to the hyperbola 3x2−y2=33 x^2-y^2=3, which are perpendicular to the line x+3y=2x+3 y=2, are
View answer and explanation
Correct option: A
y=3x±6y=3 x \pm \sqrt{6}
Option A (y=3x±6y=3 x \pm \sqrt{6}) is correct because it represents the equations of the tangents to the hyperbola that are perpendicular to the given line, calculated using the slope and the hyperbola's parameters.

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196 Easy
196 of 298
Analyzing +4 | -1
The equation of the asymptotes of the hyperbola 3x2+4y2+8xy−8x−4y−6=03{x^2} + 4{y^2} + 8xy - 8x - 4y - 6 = 0 is
View answer and explanation
Correct option: A
3x2+4y2+8xy−8x−4y−3=03{x^2} + 4{y^2} + 8xy - 8x - 4y - 3 = 0
Option A (3x2+4y2+8xy−8x−4y−3=03{x^2} + 4{y^2} + 8xy - 8x - 4y - 3 = 0) is correct because it represents a pair of straight lines and satisfies the condition for the asymptotes of the given hyperbola.
197 Easy
197 of 298
Application +4 | -1
If e1e_1 and e2e_2 are the eccentricities of the hyperbolas x225−y216=1\frac{x^2}{25}-\frac{y^2}{16}=1 and y225−x216=1\frac{y^2}{25}-\frac{x^2}{16}=1 respectively, then 1e12+1e22\frac{1}{e_1^2}+\frac{1}{e_2^2} is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because calculating the eccentricities e1e_1 and e2e_2 and then summing their reciprocals squared results in 1.
198 Easy
198 of 298
Application +4 | -1
The equation of the hyperbola whose asymptotes are 3x±5y=03 x \pm 5 y=0 and vertices are (±5,0)(\pm 5,0), is
View answer and explanation
Correct option: C
9x2−25y2=2259 x^2-25 y^2=225
Option C (9x2−25y2=2259 x^2-25 y^2=225) is the correct equation of the hyperbola because it satisfies the given asymptotes and vertices.
199 Easy
199 of 298
Analyzing +4 | -1
The combined equation of the asymptotes of the hyperbola 2x2+5xy+2y2+4x+5y=02{x^2} + 5xy + 2{y^2} + 4x + 5y = 0 is
View answer and explanation
Correct option: D
2x2+5xy+2y2+4x+5y+2=02{x^2} + 5xy + 2{y^2} + 4x + 5y + 2 = 0
Option D (2x2+5xy+2y2+4x+5y+2=02{x^2} + 5xy + 2{y^2} + 4x + 5y + 2 = 0) is correct because it represents the combined equation of the asymptotes of the given hyperbola, obtained by setting the determinant condition for a pair of straight lines to zero.
200 Easy
200 of 298
Application +4 | -1
Equation of hyperbola whose foci are (6,0) & (−6,0)(6,0) \,\&\,(-6,0) and eccentricity =2=2
View answer and explanation
Correct option: B
3x2−y2=273 x^2-y^2=27
Option B (3x2−y2=273 x^2-y^2=27) is correct because it represents the equation of a hyperbola with foci at (±6,0)(\pm 6, 0) and eccentricity 2, derived from the relationship between a, b, and e.
201 Easy
201 of 298
Application +4 | -1
If the locus of the point P‾\underline{P} represents aa hyperbola of the eccentricity ee, then the eccentricity e′e^{\prime} of the corresponding conjugate hyperbola is
View answer and explanation
Correct option: C
54\frac{5}{4}
Option C (54\frac{5}{4}) is correct because the eccentricity of the conjugate hyperbola is calculated using the relationship between the eccentricities of a hyperbola and its conjugate.
202 Easy
202 of 298
Application +4 | -1
If P\mathrm{P} is a point on the hyperbola 16x2−9y2=14416 x^2-9 y^2=144 whose foci are S1S_1 and S2S_2, then PS1∼PS2P S_1 \sim P S_2 is
View answer and explanation
Correct option: B
6
Option B (6) is correct because the difference of the distances from any point on the hyperbola to the foci is constant and equal to 2a.
203 Easy
203 of 298
Application +4 | -1
If the line y=3x+λy=3 x+\lambda touches the hyperbola 9x2−5y2=459 x^2-5 y^2=45, then λ=\lambda=
View answer and explanation
Correct option: B
±6\pm 6
Option B (±6\pm 6) is correct because it correctly uses the condition for tangency to find the value of λ\lambda.
204 Easy
204 of 298
Analyzing +4 | -1
The point P(−26, 3)P\left( { - 2\sqrt 6 ,\,\sqrt 3 } \right) lies on the hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 having eccentricity 52\frac{{\sqrt 5 }}{2}. If the tangent and normal at PP to the hyperbola intersect its conjugate axis at the point QQ and RR respectively, then QRQR is equal to
View answer and explanation
Correct option: C
636\sqrt 3
Option C (636\sqrt 3) is correct because it represents the distance QR calculated using the properties of the hyperbola, its tangent, and normal.
205 Easy
205 of 298
Application +4 | -1
Let 0<θ<π20 < \theta < \frac{\pi }{2} If the eccentricity of the hyperbola x2cos⁡2θ−y2sin⁡2θ=1\frac{{{x^2}}}{{{{\cos }^2}\theta }} - \frac{{{y^2}}}{{{{\sin }^2}\theta }} = 1 is greater than 2, then the length of its latus rectum lies in the interval
View answer and explanation
Correct option: B
(3,∞)(3,\infty )
Option B ((3,∞)(3,\infty )) is correct because the length of the latus rectum is greater than 3 when the eccentricity of the hyperbola is greater than 2.
206 Easy
206 of 298
Analyzing +4 | -1
A circle cuts rectangular hyperbola xy=1xy = 1 in the points (xr,yr),r=1,2,3,4,\left( {{x_r},{y_r}} \right),r = 1,2,3,4, then
View answer and explanation
Correct option: A
y1y2y3y4=1{y_1}{y_2}{y_3}{y_4} = 1
Option A (y1y2y3y4=1{y_1}{y_2}{y_3}{y_4} = 1) is correct because the product of the y-coordinates of the intersection points of the circle and the rectangular hyperbola xy=1 is 1.
207 Easy
207 of 298
Application +4 | -1
The equation of the hyperbola whose foci are (−2,0)(-2,0) and (2,0)(2,0) and eccentricity 2 is given by
View answer and explanation
Correct option: A
3x2−y2=33 x^2-y^2=3
Option A (3x2−y2=33 x^2-y^2=3) is correct because it correctly uses the foci and eccentricity to derive the hyperbola equation.
208 Easy
208 of 298
Application +4 | -1
The locus of the point of intersection of perpendicular tangents to the hyperbola x23−y21=1\frac{{{x^2}}}{3} - \frac{{{y^2}}}{1} = 1 is
View answer and explanation
Correct option: A
x2+y2=2{x^2} + {y^2} = 2
Option A is correct because the locus of the intersection of perpendicular tangents to a hyperbola is a circle with radius derived from the hyperbola's parameters.
209 Easy
209 of 298
Application +4 | -1
If ee and e′e' be the eccentricities of a hyperbola and its conjugate, then 1e2+1(e)2\frac{1}{{{e^2}}} + \frac{1}{{{{\left( e \right)}^2}}} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because the sum of the reciprocals of the squares of the eccentricities of a hyperbola and its conjugate is always equal to 1.
210 Easy
210 of 298
Application +4 | -1
If transverse and conjugate axes of hyperbola are equal then it’s eccentricity is
View answer and explanation
Correct option: B
2\sqrt 2
Option B (2\sqrt 2) is correct because when the transverse and conjugate axes are equal (a=b), the eccentricity of the hyperbola is calculated as 2\sqrt{2}.
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