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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Easy

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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 331-360 of 392 questions

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331 Easy
331 of 392
Application +4 | -1
If PSQ is a focal chord of the ellipse x2a2+y2b2=1,a>b\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b, then the harmonic mean of SPS P and SQS Q, is
View answer and explanation
Correct option: A
b2/ab^2 / a
Option A (b2/ab^2 / a) is correct because the harmonic mean of the segments of a focal chord is equal to the semi-latus rectum.
332 Easy
332 of 392
Application +4 | -1
An equation of the ellipse whose length of the major axis is 10 and foci are (±2,0)(\pm2, 0) is
View answer and explanation
Correct option: A
x225+y221=1\frac{x^2}{25}+\frac{y^2}{21}=1
Option A (x225+y221=1\frac{x^2}{25}+\frac{y^2}{21}=1) is correct because it satisfies the given conditions for the ellipse's major axis and foci.
333 Easy
333 of 392
Application +4 | -1
On the ellipse 2x2+3y2=12{x^2} + 3{y^2} = 1 the points at which the tangent is parallel to 4x=3y+4,4x = 3y + 4, are
View answer and explanation
Correct option: B
(−211,111)\left( { - \frac{2}{{\sqrt {11} }},\frac{1}{{\sqrt {11} }}} \right) or (211,−111)\left( {\frac{2}{{\sqrt {11} }}, - \frac{1}{{\sqrt {11} }}} \right)
Option B ((−211,111)\left( { - \frac{2}{{\sqrt {11} }},\frac{1}{{\sqrt {11} }}} \right) or (211,−111)\left( {\frac{2}{{\sqrt {11} }}, - \frac{1}{{\sqrt {11} }}} \right)) is correct because it represents the points on the ellipse where the tangent's slope matches the given line's slope.
334 Easy
334 of 392
Application +4 | -1
The equation to the ellipse, whose focus is the point (−1,1)(-1,1), whose directrix is the straight line x−y+3=0x-y+3=0 and whose eccentricity is 12\frac{1}{2} is
View answer and explanation
Correct option: A
7x2+2xy+7y2+10x−10y+7=07 x^2+2 x y+7 y^2+10 x-10 y+7=0
Option A (7x2+2xy+7y2+10x−10y+7=07 x^2+2 x y+7 y^2+10 x-10 y+7=0) is correct because it represents the equation of the ellipse derived using the focus, directrix, and eccentricity.
335 Easy
335 of 392
Application +4 | -1
The length of the axes of the conic 9x2+4y2−6x+4y+1=09{x^2} + 4{y^2} - 6x + 4y + 1 = 0, are
View answer and explanation
Correct option: C
1,  231,\;\frac{2}{3}
Option C (1, 2/3) is correct because it represents the lengths of the major and minor axes after converting the given equation to standard ellipse form.
336 Easy
336 of 392
Application +4 | -1
Chord of the ellipse x225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1 whose middle point is (12,25)(\frac{1}{2}, \frac{2}{5}) meets the minor axis at AA and major axis at B,B, length of ABAB (in units) is:
View answer and explanation
Correct option: A
415\frac{\sqrt{41}}{5}
Option A (415\frac{{\sqrt{{41}}}}{{5}}) is correct because it is the length of AB calculated using the equation of the chord with the given midpoint.
337 Easy
337 of 392
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.
338 Easy
338 of 392
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.
339 Easy
339 of 392
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.
340 Easy
340 of 392
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.
341 Easy
341 of 392
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.
342 Easy
342 of 392
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.
343 Easy
343 of 392
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.
344 Easy
344 of 392
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.
345 Easy
345 of 392
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.
346 Easy
346 of 392
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.
347 Easy
347 of 392
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.
348 Easy
348 of 392
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.
349 Easy
349 of 392
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.
350 Easy
350 of 392
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.
351 Easy
351 of 392
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.
352 Easy
352 of 392
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.
353 Easy
353 of 392
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.
354 Easy
354 of 392
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.
355 Easy
355 of 392
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.
356 Easy
356 of 392
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.
357 Easy
357 of 392
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.
358 Easy
358 of 392
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.
359 Easy
359 of 392
Application +4 | -1
The equation of the circle drawn with the two foci of x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 as the end-points of a diameter, is
View answer and explanation
Correct option: D
x2+y2=a2−b2x^2+y^2=a^2-b^2
Option D (x2+y2=a2−b2x^2+y^2=a^2-b^2) is correct because it represents the equation of the circle with the foci of the ellipse as the endpoints of a diameter.
360 Easy
360 of 392
Application +4 | -1
In the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, the equation of diameter conjugate to the diameter y=baxy=\frac{b}{a} x is
View answer and explanation
Correct option: A
y=−baxy=-\frac{b}{a} x
Option A (y=−baxy=-\frac{b}{a} x) is correct because it represents the diameter conjugate to the given diameter, satisfying the conjugate diameter condition.
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