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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Mixed difficulty

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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 511-540 of 594 questions

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511 Medium
511 of 594
Application +4 | -1
An ellipse slides between two perpendicular lines. The locus of its centre, is
View answer and explanation
Correct option: A
a circle
Option A (a circle) is correct because the locus of the center of an ellipse sliding between two perpendicular lines is a circle.
512 Easy
512 of 594
Application +4 | -1
The foci of the conic 25x2+16y2−150x=17525 x^2+16 y^2-150 x=175 are:
View answer and explanation
Correct option: C
(3,±3)(3, \pm 3)
Option C ((3,±3)(3, \pm 3)) is correct because it represents the foci of the given conic after completing the square and shifting the origin.
513 Easy
513 of 594
Application +4 | -1
An ellipse x2a2+y2b2=1(a>b)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b) passes through the point (−3,1)(-3,1) and its eccentricity is 25\sqrt{\frac{2}{5}}. Then length of latus rectum is
View answer and explanation
Correct option: A
856\frac{8}{5} \sqrt{6}
Option A (856\frac{8}{5} \sqrt{6}) is correct because it is the calculated length of the latus rectum of the ellipse based on the given conditions.
514 Easy
514 of 594
Understanding +4 | -1
The locus of the point of intersection of perpendicular tangents to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is
View answer and explanation
Correct option: C
x2+y2=a2+b2x^2+y^2=a^2+b^2
Option C (x2+y2=a2+b2x^2+y^2=a^2+b^2) is correct because the locus of the point of intersection of perpendicular tangents to an ellipse is its director circle, which has the equation x2+y2=a2+b2x^2 + y^2 = a^2 + b^2.
515 Easy
515 of 594
Application +4 | -1
The length of the latus rectum of the ellipse 9x2+4y2=19 x^2+4 y^2=1 is
View answer and explanation
Correct option: C
49\frac{4}{9}
Option C (49\frac{{4}}{{9}}) is correct because the latus rectum is calculated as 2a2b\frac{{2a^2}}{{b}} since b>ab > a.
516 Easy
516 of 594
Application +4 | -1
Number of tangents from (7,6)\left( {7,6} \right) to ellipse x216+y225=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{25}} = 1 is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the point (7,6) lies outside the ellipse, allowing two tangents to be drawn.
517 Easy
517 of 594
Analyzing +4 | -1
Tangents are drawn to the ellipse x2a2+y2b2=1(a>b)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b) and the circle x2+y2=a2x^2+y^2=a^2 at the points where a common ordinate cuts them (on the same side of the xx-axis). Then, the greatest acute angle between these tangents is given by
View answer and explanation
Correct option: A
tan⁡−1(a−b2ab)\tan ^{-1}\left(\frac{a-b}{2 \sqrt{a b}}\right)
Option A (tan⁡−1(a−b2ab)\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.
518 Easy
518 of 594
Application +4 | -1
The foci of an ellipse are (0,±6)\left( {0, \pm 6} \right) and the equations of the directrices are y=±9.y = \pm 9. The equation of the ellipse is
View answer and explanation
Correct option: D
9x2+5y2=1809{x^2} + 5{y^2} = 180
Option D (9x2+5y2=1809{x^2} + 5{y^2} = 180) is correct because it is derived from the given foci and directrices by calculating the eccentricity and using the standard equation of an ellipse.
519 Medium
519 of 594
Analyzing +4 | -1
The normal at a variable point PP on an ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 of eccentricity e meets the axes of the ellipse in QQ and RR, then the locus of the mid-point of QRQ R is a conic with eccentricity e′e^{\prime} such that
View answer and explanation
Correct option: C
e′=ee^{\prime}=e
Option C (e′=ee^{\prime}=e) is correct because the locus of the midpoint of QR is an ellipse with the same eccentricity as the original ellipse.
520 Easy
520 of 594
Application +4 | -1
The equation of the circle passing through the foci of the ellipse x216+y29=1\frac{x^2}{16}+\frac{y^2}{9}=1 and having centre at (0,3)(0,3), is
View answer and explanation
Correct option: A
x2+y2−6y−7=0x^2+y^2-6 y-7=0
Option A (x2+y2−6y−7=0x^2+y^2-6y-7=0) is correct because the equation of the circle is found using the foci of the ellipse and the given center.
521 Easy
521 of 594
Application +4 | -1
Tangent to the ellipse x232+y218=1\frac{{{x^2}}}{{32}} + \frac{{{y^2}}}{{18}} = 1 having slope −34- \frac{3}{4} meet the coordinate axes in AA and BB. Find the area of the ΔAOB,\Delta AOB, where OO is the origin
View answer and explanation
Correct option: C
24 sq unit
Option B (8 sq unit) is incorrect. Option C (24 sq unit) is correct because it's the area of the triangle formed by the tangent and the coordinate axes, calculated using the tangent's equation.
522 Easy
522 of 594
Application +4 | -1
Which one of the following is the common tangent to the ellipses, x2a2+b2+y2b2=1&x2a2+y2a2+b2=1\frac{x^2}{a^2+b^2}+\frac{y^2}{b^2}=1 \& \frac{x^2}{a^2}+\frac{y^2}{a^2+b^2}=1 ?
View answer and explanation
Correct option: B
by =ax−a4+a2b2+b4=a x-\sqrt{a^4+a^2 b^2+b^4}
Option C (5/7) is correct because the eccentricity is calculated using the distance between the foci and the sum of the distances from the origin to the foci.
523 Easy
523 of 594
Application +4 | -1
The locus of the point of intersection of tangents to an ellipse at two points whose eccentric angle differ by a constant α\alpha is x2λsec⁡2(α2)+y2μsec⁡2(α2)=1\frac{x^2}{\lambda \sec ^2\left(\frac{\alpha}{2}\right)}+\frac{y^2}{\mu \sec ^2\left(\frac{\alpha}{2}\right)}=1 then λ+μ\lambda+\mu
View answer and explanation
Correct option: A
a2+b2a^2+b^2
Option A (a2+b2a^2+b^2) is correct because the locus equation simplifies to the given form, and λ and μ are identified as a2a^2 and b2b^2 respectively.
524 Easy
524 of 594
Application +4 | -1
The tangent drawn from (α,β)\left( {\mathrm{\alpha },\mathrm{\beta }} \right) to an ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 touches the circle x2+y2=c2{x^2} + {y^2} = {c^2}, then the locus of (α,β)\left( {\mathrm{\alpha },\mathrm{\beta }} \right) is
View answer and explanation
Correct option: A
An ellipse
Option A (An ellipse) is correct because the locus of the point from which tangents to an ellipse touch a circle results in another ellipse.
525 Medium
525 of 594
Application +4 | -1
The point on the ellipse 4x2+9y2=14 x^2+9 y^2=1 where the tangent drawn is parallel to line 8x=9y8 x=9 y is
View answer and explanation
Correct option: B
(−25,15)\left(\frac{-2}{5}, \frac{1}{5}\right)
Option B ((−25,15)\left(\frac{{-2}}{{5}}, \frac{{1}}{{5}}\right)) is correct because it represents the point on the ellipse where the tangent is parallel to the given line.
526 Easy
526 of 594
Application +4 | -1
PP and QQ are the foci of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 and BB is an end of the minor axis. If PBQP B Q, is an equilateral triangle, then the eccentricity of the ellipse is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because the eccentricity is derived from the condition that PBQ forms an equilateral triangle, leading to the relationship 2ae = a.
527 Easy
527 of 594
Application +4 | -1
The tangent to the ellipse 3x2+16y2=123 x^2+16 y^2=12, at the point (1,34)\left(1, \frac{3}{4}\right) intersects the curves y2+x=0y^2+x=0 at:
View answer and explanation
Correct option: B
Exactly one point
Option B (Exactly one point) is correct because the tangent to the ellipse intersects the parabola at only one point, indicating tangency.
528 Easy
528 of 594
Analyzing +4 | -1
A circle has the same center as an ellipse and passes through the foci F1F_1 and F2F_2 of the ellipse, such that two curves intersect at four points. Let PP be any one of their point of intersection. If the major axis of the ellipse is 17 and the area of triangle PF1F2P F_1 F_2 is 30 , then the distance between the foci is
View answer and explanation
Correct option: A
13
Statement I is true, statement-II is true; statement-II is a correct explanation for statement-I because the common tangent condition leads to the equation in statement II, which then confirms the tangent equation in statement I.
529 Medium
529 of 594
Analyzing +4 | -1
If (acos⁡α,bsin⁡α),(acos⁡β,bsin⁡β)(a \cos \alpha, b \sin \alpha),(a \cos \beta, b \sin \beta) are the end points of a focal chord, then which of the following is correct?
View answer and explanation
Correct option: C
e−1e+1=tan⁡α2tan⁡β2\frac{e-1}{e+1}=\tan \frac{\alpha}{2} \tan \frac{\beta}{2}
Option C (e−1e+1=tan⁡α2tan⁡β2\frac{e-1}{e+1}=\tan \frac{\alpha}{2} \tan \frac{\beta}{2}) is correct because it represents the relationship between the eccentric angles of the endpoints of a focal chord and the eccentricity of the ellipse.
530 Easy
530 of 594
Application +4 | -1
Equation of the line passing through the points of intersection of the parabola x2=8yx^{2}=8 y and the ellipse x23+y2=1\frac{x^{2}}{3}+y^{2}=1 is :
View answer and explanation
Correct option: D
3y−1=03 y-1=0
Option D (3y−1=03 y-1=0) is correct because it represents the equation of the line passing through the intersection points of the parabola and the ellipse.
531 Easy
531 of 594
Application +4 | -1
In an ellipse, the distances between its foci is 6 and minor axis is 8. Then, its eccentricity is
View answer and explanation
Correct option: D
35\frac{3}{5}
Option D (35\frac{3}{5}) is correct because it is the eccentricity of the ellipse, calculated from the given distances between the foci and the minor axis.
532 Easy
532 of 594
Application +4 | -1
If the distance between the foci of an ellipse is equal to its minor axis, then its eccentricity is
View answer and explanation
Correct option: B
1/21 / \sqrt{2}
Option B (1/21 / \sqrt{2}) is correct because the eccentricity can be found by using the relationship between the foci, minor axis, and the eccentricity of the ellipse.
533 Easy
533 of 594
Application +4 | -1
Area bounded between 4x2+y2−8x+4y−4=04 x^{2}+y^{2}-8 x+4 y-4=0 is
View answer and explanation
Correct option: C
6π6 \pi
Option C (6π6 \pi) is correct because the equation represents an ellipse, and its area is calculated using the formula A=πabA = \pi ab, where a and b are the semi-major and semi-minor axes.
534 Easy
534 of 594
Understanding +4 | -1
If A,A′A,A' are the vertices, S,S′S,S' are the foci and Z,Z′Z,Z' are the feet of the directrices of an ellipse with centre CC, then CS,CA,CZCS,CA,CZ are in
View answer and explanation
Correct option: B
G.P
Option B (G.P) is correct because CS,CA,CZCS, CA, CZ are in geometric progression, where CA2=CS×CZCA^2 = CS \times CZ.
535 Medium
535 of 594
Application +4 | -1
If P(x,y),F1(3,0),F2(−3,0)P(x, y), F_1(3,0), F_2(-3,0) and 16x2+25y2=40016 x^2+25 y^2=400, then PF1+PF2P F_1+P F_2 equals
View answer and explanation
Correct option: C
10
Option C (10) is correct because PF1+PF2=2aPF_1 + PF_2 = 2a for any point P on the ellipse, and a=5a=5 in this case.
536 Medium
536 of 594
Analyzing +4 | -1
Locus of all such points from where the tangents drawn to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 are always inclined at 45∘45^{\circ}, is
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct because the correct locus equation differs from the provided options. The correct equation is derived using the condition for the angle between tangents.
537 Easy
537 of 594
Application +4 | -1
A point on the ellipse, 4x2+9y2=364 x^{2}+9 y^{2}=36, where the normal is parallel to the line, 4x−2y−5=04 x-2 y-5=0, is :
View answer and explanation
Correct option: C
(−95,85)\left(-\frac{9}{5}, \frac{8}{5}\right)
Option C ((−95,85)\left(-\frac{9}{5}, \frac{8}{5}\right)) is correct because it represents the point on the ellipse where the normal is parallel to the given line.
538 Medium
538 of 594
Analyzing +4 | -1
The locus of the mid-points of the chords of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 whose poles are on the auxiliary circle on the tangents at the extremities of which intersect on the auxiliary circle, is
View answer and explanation
Correct option: D
x2+y2=a2(x2a2+y2b2)2x^2+y^2=a^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)^2
Option D (x2+y2=a2(x2a2+y2b2)2x^2+y^2=a^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)^2) is correct because it represents the locus of mid-points of chords whose poles lie on the auxiliary circle.
539 Medium
539 of 594
Application +4 | -1
The equation of the ellipse whose axes are coincident with the coordinate axes and which touches the straight line 3x−2y−20=03x-2y-20 = 0 and x+6y−20=0x+6 y-20=0, is
View answer and explanation
Correct option: C
x240+y210=1\frac{x^2}{40}+\frac{y^2}{10}=1
Option C (x240+y210=1\frac{x^2}{40}+ \frac{y^2}{10}=1) is correct because it represents the equation of the ellipse that touches both given lines, found by using the tangent condition.
540 Easy
540 of 594
Application +4 | -1
If (3)bx+ay=2ab\left( {\sqrt 3 } \right)bx + ay = 2ab touches the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1, then eccentric angle ϕ\phi is
View answer and explanation
Correct option: A
π6\frac{\pi }{6}
Option A (π6\frac{\pi }{6}) is correct because it is the eccentric angle that satisfies the given equation of the tangent to the ellipse.
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