If a variable tangent of the circle $x^2+y^2=1$ intersect the ellipse $x^2+2 y^2=4$ at $\mathrm{P}$ and $\mathrm{Q}$ then the locus of the points of intersection of the tangents are $\mathrm{P}$ and $\mathrm{Q}$ is

A parabola with focus as $(2,3)$

An ellipse with eccentricity $\frac{\sqrt{3}}{4}$

Back to ELLIPSE

Your Progress

0 / 594 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →

Showing 301-330 of 594 questions

Preparing practice and saved progress…

301 Easy

301 of 594

Application +4 | -1

Eccentricity of the ellipse

{x^2} + 2{y^2} - 2x + 3y + 2 = 0

View answer and explanation

Correct option: A

\frac{1}{{\sqrt 2 }}

Option A

\frac{1}{{\sqrt 2 }}

is correct because the eccentricity of the ellipse is found by rewriting the equation in standard form and using the formula

e = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}}

302 Medium

302 of 594

Application +4 | -1

P\left( {\alpha ,\beta } \right)

is a point on the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1

with foci

S

and

S'

and eccentricity

e

, then area of

\mathrm{\Delta }SPS'

View answer and explanation

Correct option: D

be\sqrt {{a^2} - {\alpha ^2}}

Option D (

be\sqrt {{a^2} - {\alpha ^2}}

) is correct because it represents the area of the triangle formed by the foci and a point on the ellipse, derived using the properties of the ellipse.

303 Easy

303 of 594

Application +4 | -1

If the line

y=m x+c

touches the ellipse

\frac{x^2}{b^2}+\frac{y^2}{a^2}=1

, then find the value of

c

View answer and explanation

Correct option: A

\pm \sqrt{b^2 m^2+a^2}

Option A (

\pm \sqrt{{b^2 m^2+a^2}}

) is correct because, for the ellipse

\frac{{x^2}}{{b^2}} + \frac{{y^2}}{{a^2}} = 1

, the condition for tangency of the line y = mx + c is

c = \pm \sqrt{{a^2 + b^2 m^2}}

due to the interchange of a and b.

304 Medium

304 of 594

Application +4 | -1

The locus of mid-points of a focal chord of the ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

, is

View answer and explanation

Correct option: A

\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{e x}{a}

Option A (

\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{e x}{a}

) is correct because it represents the locus of the midpoints of the focal chords of the ellipse.

305 Medium

305 of 594

Analyzing +4 | -1

The area of the triangle inscribed in an ellipse is

2 a b \sin \left(\frac{\beta-\gamma}{2}\right) \sin \left(\frac{\gamma-\alpha}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)

, where

\alpha, \beta

\gamma

are the eccentric angles of the vertices, then the condition that the area of a triangle inscribed in an ellipse may be maximum, is

View answer and explanation

Correct option: B

\alpha, \alpha+\frac{2 \pi}{3}, \alpha+\frac{4 \pi}{3}

Option B (

\alpha, \alpha+\frac{2 \pi}{3}, \alpha+\frac{4 \pi}{3}

) is correct because the area of the triangle is maximum when the eccentric angles are equally spaced around the ellipse, which occurs when they are

\alpha, \alpha+\frac{2 \pi}{3}, \alpha+\frac{4 \pi}{3}

306 Easy

306 of 594

Application +4 | -1

Equation of the tangents to the ellipse

\frac{x^2}{9}+\frac{y^2}{16}=1

which are parallel to the line

x+y+1=0

, is

View answer and explanation

Correct option: A

x+y+5=0

Option A (

x+y+5=0

) is correct because it satisfies the condition for tangency to the ellipse and is parallel to the given line.

307 Hard

307 of 594

Analyzing +4 | -1

If circumcentre of an equilateral triangle inscribed in

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

, with vertices having eccentric angles

\alpha, \beta, \gamma

, respectively is

\left(x_1, y_1\right)

then

\Sigma \cos \alpha \cos \beta+\Sigma \sin \alpha \sin \beta=

View answer and explanation

Correct option: D

\frac{9 x_1{ }^2}{2 a^2}+\frac{9 y_1^2}{2 b^2}-\frac{3}{2}

Option D (

\frac{9 x_1{ }^2}{2 a^2}+\frac{9 y_1^2}{2 b^2}-\frac{3}{2}

) is correct because it represents the correct relationship between the coordinates of the circumcenter and the eccentric angles of the vertices of the equilateral triangle.

308 Easy

308 of 594

Application +4 | -1

The curve represented by

x=2(\cos t+\sin t)

y=5(\cos t-\sin t)

, is

View answer and explanation

Correct option: C

an ellipse

Option C (an ellipse) is correct because the given parametric equations can be manipulated to form the equation of an ellipse.

309 Medium

309 of 594

Analyzing +4 | -1

p

and

q

are the segments of a focal chord of an ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,

then

View answer and explanation

Correct option: B

{b^2}\left( {p + q} \right) = 2apq

Option B is correct because it directly applies the harmonic mean relationship for focal chords of an ellipse.

310 Medium

310 of 594

Application +4 | -1

If the distance between the foci of an ellipse is 8 and distance between its directrix is 10, then length of its latus rectum is

View answer and explanation

Correct option: C

\frac{4}{\sqrt{5}}

Option C (

\frac{4}{\sqrt{5}}

) is correct because after finding the eccentricity and semi-major axis, the latus rectum can be calculated using the formula

\frac{2b^2}{a}

311 Medium

311 of 594

Application +4 | -1

The coordinates of the middle point of the chord intercepted on the line

2 x-y+3=0

by the ellipse

\frac{x^2}{10}+\frac{y^2}{6}=1

, are

View answer and explanation

Correct option: A

\left(\frac{-30}{23}, \frac{9}{23}\right)

Option A (

\left(\frac{-30}{23}, \frac{9}{23}\right)

) is correct because it represents the midpoint of the chord intercepted on the line by the ellipse.

312 Easy

312 of 594

Analyzing +4 | -1

Let

S

and

S^{\prime}

be two foci of the ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

. If a circle described on

SS'

as diameter intersects the ellipse in real and distinct points, then the eccentricity

e

of the ellipse satisfies

View answer and explanation

Correct option: B

e \in(1 / \sqrt{2}, 1)

Option A (

\tan ^{-1}\left(\frac{a-b}{2 \sqrt{a b}}\right)

) is correct because it represents the greatest acute angle between the tangents to the ellipse and the circle at the points where a common ordinate cuts them.

313 Easy

313 of 594

Application +4 | -1

The distances from the foci of point

P\left( {{x_1},{y_1}} \right)

on the ellipse

\frac{{{x^2}}}{9} + \frac{{{y^2}}}{{25}} = 1

are

View answer and explanation

Correct option: C

5 \pm \frac{4}{5}{y_1}

Option C (

5 \pm \frac{4}{5}{y_1}

) is correct because it represents the focal distances of a point on the ellipse, considering that the major axis is along the y-axis.

314 Easy

314 of 594

Application +4 | -1

If two tangents drawn to the ellipse

x^2+4 y^2=4

intersect perpendicularly at

P

, then the locus of

P

View answer and explanation

Correct option: B

x^2+y^2=5

Option B (

x^2+y^2=5

) is correct because the locus of P is the director circle of the ellipse, and its equation is

x^2 + y^2 = a^2 + b^2

315 Easy

315 of 594

Application +4 | -1

Let

P

be a point in the first quadrant lying on the ellipse

9 x^2+16 y^2=144

, such that the tangent at

P

to the ellipse is inclined at an angle of

135^{\circ}

to the positive direction of

x

-axis. Then the coordinates of

P

are

View answer and explanation

Correct option: A

\left(\frac{16}{5}, \frac{9}{5}\right)

Option A ((16/5, 9/5)) is correct because it satisfies the ellipse equation and the tangent at that point has a slope of -1.

Explore Other Chapters in MATHEMATICS

View All →

SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

316 Easy

316 of 594

Application +4 | -1

The angle between the pair of tangents drawn to the ellipse

3{x^2} + 2{y^2} = 5

from the point

\left( {1,2} \right)

View answer and explanation

Correct option: C

{\tan ^{ - 1}}\left( {\frac{{12}}{{\sqrt 5 }}} \right)

Option C (

{\tan ^{ - 1}}(\frac{{12}}{{\sqrt 5 }}})

) is correct because it is the angle between the pair of tangents calculated using the joint equation of tangents.

317 Easy

317 of 594

Application +4 | -1

The distance of the point

(\sqrt{6} \cos \theta, \sqrt{2} \sin \theta)

on the ellipse from the centre of the ellipse is 2 if

\theta=

View answer and explanation

Correct option: B

\pi/4

Option B (

\pi/4

) is correct because substituting the point into the distance formula and solving for

\theta

yields

\pi/4

318 Easy

318 of 594

Application +4 | -1

If the coordinates of the centre, a focus and adjacent vertex are

\left( {2, - 3} \right),\left( {3, - 3} \right)

and

\left( {4, - 3} \right)

respectively, then the equation of the ellipse is

View answer and explanation

Correct option: A

\frac{{{{\left( {x - 2} \right)}^2}}}{4} + \frac{{{{\left( {y + 3} \right)}^2}}}{3} = 1

Option A (

\frac{{{{\left( {x - 2} \right)}^2}}}{4} + \frac{{{{\left( {y + 3} \right)}^2}}}{3} = 1

) is correct because it correctly represents an ellipse with the given center, focus, and vertex.

319 Medium

319 of 594

Analyzing +4 | -1

If a variable tangent of the circle

x^2+y^2=1

intersect the ellipse

x^2+2 y^2=4

\mathrm{P}

and

\mathrm{Q}

then the locus of the points of intersection of the tangents are

\mathrm{P}

and

\mathrm{Q}

View answer and explanation

Correct option: D

An ellipse with length of latus rectum is 2 units

The correct answer is D. The locus of the points of intersection of the tangents is an ellipse, and the length of its latus rectum is calculated to be 2 units.

320 Medium

320 of 594

Application +4 | -1

{x^2} + 9{y^2} - 4x + 3 = 0,x,y \in R

, then

x

and

y

respectively lie in the intervals

View answer and explanation

Correct option: C

\mathrm{[1,3]and[1,3]}

The correct answer is

\mathrm{[1,3]and[1,3]}

because the equation represents an ellipse, and the ranges for x and y can be determined from the standard form of the ellipse equation.

321 Easy

321 of 594

Analyzing +4 | -1

x-2 y+4=0

is a common tangent to

y^2=4 x

and

\frac{x^2}{4}+\frac{y^2}{b^2}=1

. Then, the value of

b

and the other common tangent are given by

View answer and explanation

Correct option: A

b=\sqrt{3} ; x+2 y+4=0

Option A (

b=\sqrt{{3}} ; x+2 y+4=0

) is correct because the given line is tangent to both the parabola and the ellipse, and the other common tangent is found using symmetry.

322 Easy

322 of 594

Application +4 | -1

For the ellipse

25 x^2+9 y^2-150 x-90 y+225=0

, the eccentricity is

View answer and explanation

Correct option: C

4 / 5

Option C (4 / 5) is correct because, after rewriting the equation in standard form, the eccentricity is calculated using the formula

e = \sqrt{{1 - \frac{{a^2}}{{b^2}}}}

since b > a.

323 Easy

323 of 594

Application +4 | -1

Locus of the point of intersection of tangents at the end points of focal chord is

View answer and explanation

Correct option: A

x=\frac{a}{e^\prime}

a > b

Option A (

x=\frac{a}{e^{\prime}}

, if

a > b

) is correct because the locus of the intersection of tangents at the endpoints of a focal chord is the directrix of the ellipse.

324 Easy

324 of 594

Application +4 | -1

x=6 y+8

be a tangent to the ellipse in standard form whose minor axis is

\frac{4}{\sqrt{3}}

, then its eccentricity is

View answer and explanation

Correct option: B

\sqrt{\frac{11}{12}}

Option B (

\sqrt{\frac{11}{12}}

) is correct because it is the eccentricity calculated using the tangent condition and the given minor axis length.

325 Easy

325 of 594

Application +4 | -1

The eccentricity of the ellipse represented by the equation

25\;{x^2} + 16\;{y^2} - 150\;x - 175 = 0

View answer and explanation

Correct option: B

3/5

Option B is correct because the eccentricity of the ellipse is calculated after rewriting the equation in standard form and using the relationship between a, b, and e.

326 Easy

326 of 594

Application +4 | -1

The eccentricity of the ellipse

4 x^2+y^2-8 x+2 y+1=0

View answer and explanation

Correct option: B

\sqrt{3} / 2

Option B (

\sqrt{3} / 2

) is correct because the eccentricity is calculated using the formula

e = \sqrt{{1 - \frac{{a^2}}{{b^2}}}}

after rewriting the equation in standard ellipse form.

327 Easy

327 of 594

Application +4 | -1

The equation of the ellipse, whose one of the vertices is

(0,7)

and the corresponding directrix is

y=12

, is

View answer and explanation

Correct option: B

144 x^2+95 y^2=4655

Option B (

144 x^2+95 y^2=4655

) is the correct equation of the ellipse because it satisfies the given vertex and directrix conditions.

328 Hard

328 of 594

Application +4 | -1

The length of the smallest intercept made by the coordinate axes on any tangent to the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1

, is

View answer and explanation

Correct option: A

a + b

Option A (

a + b

) is correct because it represents the minimum length of the intercept made by the coordinate axes on any tangent to the ellipse.

329 Easy

329 of 594

Analyzing +4 | -1

Let a line

L

pass through the point intersection of the lines

b x+10 y-8=0

and

2 x-3 y=0

b \in R-\left\{\frac{4}{3}\right\}

. If the line

L

also passes through the point

(1,1)

and touches the circle

17\left(x^2+y^2\right)=16

, then the eccentricity of the ellipse

\frac{x^2}{5}+\frac{y^2}{5}=1

View answer and explanation

Correct option: B

\sqrt{\frac{3}{5}}

Option D (

\sqrt{\frac{3}{5}}

) is correct because after finding the equation of the line L and using the condition that it touches the circle, we can find the value of b and then calculate the eccentricity.

330 Easy

330 of 594

Remembering +4 | -1

P

is any point on the ellipse

\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{16}} = 1

and

S

and

S'

are the foci, then

PS + PS'

is equal to

View answer and explanation

Correct option: D

Option D (12) is correct because the sum of the focal distances of any point on the ellipse is equal to 2a, where a = 6 in this case.

Revision notes and key takeaways

Review ELLIPSE in your textbook or class notes, then use this set to check your understanding.

Showing 301-330 of 594 questions

Your Progress

Ellipse practice · Mixed difficulty

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice