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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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541 Medium
541 of 594
Analyzing +4 | -1
The point, at shortest distance from the line x+y=7x + y = 7 and lying on an ellipse x2+2y2=6{x^2} + 2{y^2} = 6, has coordinates
View answer and explanation
Correct option: C
(2,1)\left( {2,1} \right)
Option C ((2,1)) is correct because the tangent to the ellipse at this point is parallel to the given line x + y = 7.
542 Medium
542 of 594
Application +4 | -1
If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is 32\frac{3}{2} units, then its eccentricity is?
View answer and explanation
Correct option: D
13\frac{1}{3}
Option D (13\frac{{1}}{{3}}) is correct because it is derived from the given latus rectum length and the distance between a focus and its nearest vertex.
543 Easy
543 of 594
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.
544 Easy
544 of 594
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.
545 Easy
545 of 594
Application +4 | -1
The foci of the ellipse (x−3)236+(y+2)216=1\frac{(x-3)^2}{36}+\frac{(y+2)^2}{16}=1, are
View answer and explanation
Correct option: B
(3±25,−2)(3 \pm 2 \sqrt{5},-2)
Option B ((3±25,−2)(3 \pm 2 \sqrt{5},-2)) is correct because it represents the foci of the ellipse after shifting the origin and calculating the eccentricity.
546 Easy
546 of 594
Application +4 | -1
Length of the major axis of the ellipse 9x2+7y2=639 x^2+7 y^2=63, is
View answer and explanation
Correct option: C
6
Option C (6) is correct because the length of the major axis is 2b, where b is the semi-major axis, and in this case, b=3.
547 Easy
547 of 594
Application +4 | -1
The locus of the point of intersection of the perpendicular tangents to the ellipse x29+y24=1\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1 is
View answer and explanation
Correct option: C
x2+y2=13{x^2} + {y^2} = 13
Option C (x2+y2=13{x^2} + {y^2} = 13) is correct because the locus of the intersection of perpendicular tangents to an ellipse is its director circle, and the equation is x2+y2=a2+b2{x^2} + {y^2} = {a^2} + {b^2}.
548 Medium
548 of 594
Application +4 | -1
The sum of the squares of the perpendiculars on any tangent to the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 from two points on the minor axis each at a distance a2−b2\sqrt {{a^2} - {b^2}} from the centre is
View answer and explanation
Correct option: A
2a22{a^2}
Option A (2a22{a^2}) is correct because the sum of the squares of the perpendiculars from the specified points on the minor axis to any tangent of the ellipse simplifies to 2a22a^2.
549 Easy
549 of 594
Application +4 | -1
Let the eccentricity of an ellipse x2a2+y2b2=1, a>b,\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\,a > b, be 14.\frac{1}{4}. If this ellipse passes through the point (−425,3),\left( { - 4\sqrt {\frac{2}{5}} ,3} \right), then a2+b2{a^2} + {b^2} is equal to:
View answer and explanation
Correct option: B
31
Option B (31) is correct because by substituting the given point and eccentricity into the ellipse equation, we can solve for a² + b².
550 Medium
550 of 594
Application +4 | -1
The eccentricity of the ellipse, whose latus rectum is equal to the distance between two focus points, is
View answer and explanation
Correct option: B
5−12\frac{\sqrt{5}-1}{2}
Option B (5−12\frac{{\sqrt{{5}}-1}}{{2}}) is correct because it is the positive solution to the equation derived from the given condition relating latus rectum and focal distance.
551 Easy
551 of 594
Application +4 | -1
If the length of the major axis of an ellipse is 178\frac{{17}}{8} times the length of the minor axis, then the eccentricity of the ellipse is
View answer and explanation
Correct option: B
1517\frac{{15}}{{17}}
Option B (1517\frac{{15}}{{17}}) is correct because the eccentricity is calculated using the relationship between the major and minor axes and the formula b2=a2(1−e2)b^2 = a^2(1 - e^2).
552 Easy
552 of 594
Application +4 | -1
The locus of the point of intersection of tangents to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 at the points whose eccentric angles differ by π/2,\pi / 2, is
View answer and explanation
Correct option: D
x2a2+y2b2=2\frac{x^2}{a^2}+\frac{y^2}{b^2}=2
Option D (x2a2+y2b2=2\frac{x^2}{a^2}+\frac{y^2}{b^2}=2) is correct because it represents the locus of the intersection point of tangents to the ellipse where the eccentric angles differ by π/2\pi/2.
553 Easy
553 of 594
Application +4 | -1
The equation of the ellipse whose distance between the foci is equal to 8 and distance between the directrices is 18 , is
View answer and explanation
Correct option: D
5x2+9y2=1805 x^2+9 y^2=180
Option D (5x2+9y2=1805 x^2+9 y^2=180) is correct because using the given information about the foci and directrices, we can derive the equation of the ellipse as 5x2+9y2=1805x^2 + 9y^2 = 180.
554 Easy
554 of 594
Application +4 | -1
If the eccentricity of the ellipse x2a2+2+y2a2+5=1\frac{x^2}{a^2+2}+\frac{y^2}{a^2+5}=1 be 13\frac{1}{\sqrt{3}}, then length of latus rectum of ellipse is
View answer and explanation
Correct option: A
4
Option A (4) is correct because the length of the latus rectum of the ellipse is calculated by first finding the values of a and b using the given eccentricity and then applying the formula for the latus rectum.
555 Easy
555 of 594
Application +4 | -1
The equation 14x2−4xy+11y2−44x−58y+71=014 x^2-4 x y+11 y^2-44 x-58 y+71=0 represents
View answer and explanation
Correct option: B
An ellipse
Option B (An ellipse) is correct because the discriminant h2−abh^2 - ab is less than 0, indicating an ellipse.
556 Medium
556 of 594
Application +4 | -1
If pp is the length of the perpendicular from a facus upon the tangent at any point pp of the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 and rr is the distance of pp from the focus, then 2ar−b2p2\frac{{2a}}{r} - \frac{{{b^2}}}{{{p^2}}} is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because, after substituting and simplifying, the expression equals 1.
557 Easy
557 of 594
Analyzing +4 | -1
Consider ellipses Ek:kx2+k2y2=1,k=1,2,…,20E_k: k x^2+k^2 y^2=1, k=1,2, \ldots, 20.

Let CkC_k be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse EkE_k. If rkr_k is the radius of the circle CkC_k, then the value of ∑k=1201rk2\sum\limits_{{k=1}}^{20} \frac{1}{r_k^2} is
View answer and explanation
Correct option: A
3080
Option A is correct. By finding the radius of the circle that touches the chords and summing the reciprocals of the squares of the radii, we arrive at 3080.
558 Easy
558 of 594
Application +4 | -1
The equation of the tangent at the point (14,14)\left(\frac{1}{4}, \frac{1}{4}\right) of the ellipse x24+y212=1\frac{x^2}{4}+\frac{y^2}{12}=1 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the point (1/4, 1/4) does not lie on the ellipse, so a tangent cannot be defined at that point.
559 Easy
559 of 594
Application +4 | -1
The eccentricity of an ellipse is 2/32 / 3, its latus rectum is 5 and its centre is (0,0)(0,0). The equation of the ellipse is
View answer and explanation
Correct option: B
4x281+4y245=1\frac{4 x^2}{81}+\frac{4 y^2}{45}=1
Option C (y2=2xy^2=2 x) is correct because the parabola passes through the intersection points of the line and the circle.
560 Easy
560 of 594
Application +4 | -1
Let x=4x=4 be a directrix to an ellipse whose centre is at the origin and eccentricity is 12\frac{1}{2}.If P(1,β),β>0P(1, \beta), \beta>0 is a point on the ellipse, then the equation of the normal to it at PP, is
View answer and explanation
Correct option: B
4x−2y=14 x-2 y=1
Option A (4x−2y=14 x-2 y=1) is correct because after finding the equation of the ellipse and the point P, the equation of the normal is calculated correctly.
561 Medium
561 of 594
Application +4 | -1
Let EE be the ellipse x28+y24=1\frac{{{x^2}}}{8} + \frac{{{y^2}}}{4} = 1 and CC be the circle x2+y2=9{x^2} + {y^2} = 9. Let PP and QQ be the points (1,2)\left( {1,2} \right) and (2,1)\left( {2,1} \right) respectively. Then
View answer and explanation
Correct option: D
PP lies inside CC but outside EE
Option D (PP lies inside CC but outside EE) is correct because substituting the coordinates of P into the equations of the circle and ellipse shows that P is inside the circle and outside the ellipse.
562 Medium
562 of 594
Understanding +4 | -1
The equation of a diameter conjugate to a diameter y=baxy = \frac{b}{a}x of the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 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 conjugate diameters of an ellipse satisfy the condition m1m2=−b2a2{m_1}{m_2} = - \frac{{{b^2}}}{{{a^2}}}.
563 Medium
563 of 594
Application +4 | -1
The locus of the point of intersection of tangents to the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 which meet at right angle is
View answer and explanation
Correct option: A
A circle
Option A A circle is correct because the locus of the point of intersection of perpendicular tangents to an ellipse is its director circle, which is a circle.
564 Medium
564 of 594
Understanding +4 | -1
In an ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of the ellipse to the point of contact meet on
View answer and explanation
Correct option: B
Corresponding directrix
Option B (Corresponding directrix) is correct because the intersection point lies on the directrix of the ellipse, as derived from the tangent and perpendicular equations.
565 Easy
565 of 594
Application +4 | -1
The equation of the tangents to the ellipse 4x2+3y2=54 x^2+3 y^2=5 which are parallel to the line y=3x+7y=3 x+7 are
View answer and explanation
Correct option: B
y=3x±15512y=3 x \pm \sqrt{\frac{155}{12}}
Option B (y=3x±15512y=3 x \pm \sqrt{\frac{{155}}{{12}}}) is correct because it represents the tangent lines to the ellipse with a slope of 3.
566 Medium
566 of 594
Application +4 | -1
If CC is the centre of the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 and the normal at an end of a latusrectum cuts the major axis in G,G, then CG=CG =
View answer and explanation
Correct option: C
ae2a{e^2}
Option C is correct because the distance CG is calculated using the equation of the normal at the end of the latus rectum and the eccentricity of the ellipse.
567 Easy
567 of 594
Application +4 | -1
Equation of the normal to the ellipse 4(x−1)2+9(y−2)2=364{\left( {x - 1} \right)^2} + 9{\left( {y - 2} \right)^2} = 36, which is parallel to the line 3x−y=13x - y = 1, is
View answer and explanation
Correct option: D
3x−y=5(5+1)3x - y = \sqrt 5 \left( {\sqrt 5 + 1} \right)
Option D (3x−y=5(5+1)3x - y = \sqrt 5 (\sqrt 5 + 1)) is correct because it represents the equation of the normal to the given ellipse that is parallel to the line 3x−y=13x - y = 1.
568 Medium
568 of 594
Application +4 | -1
The locus of the middle points of the portions of the tangents of the ellipse x2/a2+y2/b2=1x^2 / a^2+y^2 / b^2=1 included between the axis is the curve.
View answer and explanation
Correct option: B
a2/x2+b2/y2=4a^2 / x^2+b^2 / y^2=4
Option B (a2/x2+b2/y2=4a^2 / x^2+b^2 / y^2=4) is correct because it represents the locus of the midpoints of the tangent segments of the ellipse intercepted by the axes.
569 Medium
569 of 594
Application +4 | -1
The line x=at2x = a{t^2} meets the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 in the real points, if
View answer and explanation
Correct option: B
∣t∣≤1\left| t \right| \le 1
Option B (∣t∣≤1\left| t \right| \le 1) is correct because it ensures that the value inside the square root for calculating y is non-negative, resulting in real points of intersection.
570 Medium
570 of 594
Application +4 | -1
Tangents are drawn from the point on the line x−y−5=0x - y - 5 = 0 to x2+4y2=4{x^2} + 4{y^2} = 4, then all the chords of contact pass through a fixed point, whose coordinates are
View answer and explanation
Correct option: D
(45,  −15)\left( {\frac{4}{5},\; - \frac{1}{5}} \right)
Option D ((45,  −15)\left( {\frac{4}{5},\; - \frac{1}{5}} \right)) is correct because the chord of contact passes through a fixed point which is the intersection of two lines derived from the equation of the chord of contact.
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