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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 391-420 of 594 questions

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391 Easy
391 of 594
Application +4 | -1
The product of the perpendiculars drawn from the two foci of an ellipse to the tangent at any point of the ellipse, is
View answer and explanation
Correct option: B
b2b^2
Option B (b2b^2) is correct because the product of perpendiculars from foci to any tangent of an ellipse equals the square of the semi-minor axis.
392 Easy
392 of 594
Application +4 | -1
The area (in square units) of the quadrilateral formed by the tangents at the end-points of latusrecta to the ellipse x29+y25=1\frac{x^2}{9}+\frac{y^2}{5}=1, is
View answer and explanation
Correct option: B
27
Option B (27) is correct because it represents the area of the quadrilateral formed by the tangents at the endpoints of the latus recta of the ellipse.
393 Medium
393 of 594
Application +4 | -1
If the length of the latusrectum of the ellipse x2tan⁡2θ+y2sec⁡2θ=1{x^2}{\tan ^2}\theta + {y^2}{\sec ^2}\theta = 1 is 12\frac{1}{2}, then θ=\theta =
View answer and explanation
Correct option: A
π/12,5π/12\pi /12,5\pi /12
Option A (π/12,5π/12\pi /12,5\pi /12) is correct because these are the values of θ\theta that satisfy the given condition for the length of the latus rectum.
394 Easy
394 of 594
Application +4 | -1
If xx and yy are real numbers such that (x−4)24+y29=1\frac{(x-4)^2}{4}+\frac{y^2}{9}=1, then the largest possible value of x24+y29\frac{x^2}{4}+\frac{y^2}{9} is
View answer and explanation
Correct option: A
9
Option A (9) is correct because by parameterizing the ellipse and maximizing the given expression, the largest possible value is found to be 9.
395 Easy
395 of 594
Application +4 | -1
The sum of the squares of the perpendicular 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 simplifies to 2a22a^2 after algebraic manipulation and using the ellipse properties.
396 Easy
396 of 594
Application +4 | -1
The area of the triangle formed by three points on the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 whose eccentric angles are α,β\alpha ,\beta and γ\gamma is
View answer and explanation
Correct option: C
2absin⁡α−β2sin⁡β−γ2sin⁡γ−α22ab\sin \frac{{\alpha - \beta }}{2}\sin \frac{{\beta - \gamma }}{2}\sin \frac{{\gamma - \alpha }}{2}
Option C (2absin⁡α−β2sin⁡β−γ2sin⁡γ−α22ab\sin \frac{{\alpha - \beta }}{2}\sin \frac{{\beta - \gamma }}{2}\sin \frac{{\gamma - \alpha }}{2}) is correct because it represents the area of the triangle formed by the given points on the ellipse using eccentric angles.
397 Easy
397 of 594
Understanding +4 | -1
The equation x22−r+y2r−5+1=0\frac{x^2}{2-r}+\frac{y^2}{r-5}+1=0 represent the ellipse if
View answer and explanation
Correct option: B
2<r<52 < r < 5
Option D (r∈(2,5)r \in (2,5)) is correct because for the equation to represent an ellipse, both denominators must be positive.
398 Easy
398 of 594
Application +4 | -1
In an ellipse the distance between the foci is 8 and the distance between the directrices is 25. The length of major axis is
View answer and explanation
Correct option: A
10210\sqrt 2
The correct answer is A (10210\sqrt 2) because it represents the length of the major axis of the ellipse, calculated using the given distances between the foci and directrices.
399 Easy
399 of 594
Understanding +4 | -1
The sum of the focal distances of any point on the conic x225+y216=1\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1 is
View answer and explanation
Correct option: A
10
Option C ((aet+e−t2,bet−e−t2)\left( {a\frac{{{e^t} + {e^{ - t}}}}{2},b\frac{{{e^t} - {e^{ - t}}}}{2}} \right)) is correct because these parametric coordinates satisfy the equation of the hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1.
400 Medium
400 of 594
Analyzing +4 | -1
If QQ is the point on the auxiliary circle corresponding to a point PP on an ellipse.
Then, the normals at PP and QQ meet on
View answer and explanation
Correct option: A
A fixed circle
Option A (A fixed circle) is correct because the normals at P and Q meet on a fixed circle.
401 Medium
401 of 594
Analyzing +4 | -1
A tangent at any point to the ellipse 4x2+9y2=364{x^2} + 9{y^2} = 36 is cut by the tangent at the extremities of the major axis at TT and T′T'. The circle on TT′TT' as diameter passes through the point
View answer and explanation
Correct option: B
(5,0)\left( {\sqrt 5 ,0} \right)
Option B ((5,0)\left( {\sqrt 5 ,0} \right)) is correct because the circle on TT' as diameter passes through this point, as derived from the equation of the circle.
402 Medium
402 of 594
Application +4 | -1
The eccentricity of the ellipse 4x2+9y2=364 x^2+9 y^2=36, is
View answer and explanation
Correct option: C
53\frac{\sqrt{5}}{3}
Option C (53\frac{{\sqrt{{5}}}}{{3}}) is correct because the eccentricity is calculated using the formula e=1−b2a2e = \sqrt{{1 - \frac{{b^2}}{{a^2}}}} after converting the equation to standard form.
403 Easy
403 of 594
Application +4 | -1
The value of kk, if (1,2),\left( {1,2} \right), (k,−1)\left( {k, - 1} \right) are conjugate points with respect to the ellipse 2x2+3y2=6,2{x^2} + 3{y^2} = 6, is
View answer and explanation
Correct option: C
6
The correct answer is 6 because it satisfies the condition for conjugate points with respect to the given ellipse.
404 Easy
404 of 594
Application +4 | -1
The foci of the ellipse, 25(x+1)2+9(y+2)2=22525{\left( {x + 1} \right)^2} + 9{\left( {y + 2} \right)^2} = 225 are at,
View answer and explanation
Correct option: A
(−1,2)\left( { - 1,2} \right) and (−1,−6)\left( { - 1, - 6} \right)
The correct answer is A ((-1,2) and (-1,-6)) because it correctly identifies the foci of the ellipse after translating the standard ellipse equation.
405 Medium
405 of 594
Analyzing +4 | -1
Let the tangent and normal at the point (33,1)(3 \sqrt{3}, 1) on the ellipse x236+y24=1\frac{x^{2}}{36}+\frac{y^{2}}{4}=1 meet the yy-axis at the points AA and BB respectively. Let the circle CC be drawn taking ABA B as a diameter and the line x=25x=2 \sqrt{5} intersect CC at the points PP and QQ. If the tangents at the points PP and QQ on the circle intersect at the point (α,β)(\alpha, \beta), then α2−β2\alpha^{2}-\beta^{2} is equal to
View answer and explanation
Correct option: C
3045\frac{304}{5}
The correct answer is C (3045\frac{304}{5}) because it is the value of α2−β2\alpha^2 - \beta^2 after finding the intersection point of tangents to the circle constructed with the segment AB as diameter.
406 Easy
406 of 594
Application +4 | -1
If e1{e_1} and e2{e_2} are the eccentricities of the ellipse, x218+y24=1\frac{{{x^2}}}{{18}} + \frac{{{y^2}}}{4} = 1 and the hyperbola, x29−y24=1\frac{{{x^2}}}{9} - \frac{{{y^2}}}{4} = 1 respectively and (e1,e2)\left( {{e_1},{e_2}} \right) is a point on the ellipse, 15x2+3y2=k15{x^2} + 3{y^2} = k, then kk is equal to
View answer and explanation
Correct option: D
16
Option D (16) is correct because after calculating the eccentricities of the ellipse and hyperbola, substituting them into the given ellipse equation yields k=16.
407 Easy
407 of 594
Application +4 | -1
If e1{e_1} and e2{e_2} are respectively the eccentricities of the ellipse x218+y24=1\frac{{{x^2}}}{{18}} + \frac{{{y^2}}}{4} = 1 and the hyperbola x29−y24=1,\frac{{{x^2}}}{9} - \frac{{{y^2}}}{4} = 1, then the relation between e1{e_1} and e2{e_2} is
View answer and explanation
Correct option: C
2e12+e22=32e_1^2 + e_2^2 = 3
Option C (2e12+e22=32e_1^2 + e_2^2 = 3) is correct because it represents the correct relationship between the eccentricities of the given ellipse and hyperbola.
408 Easy
408 of 594
Application +4 | -1
If the polar with respect to y2=4ax{y^2} = 4ax touches the ellipse x2α2+y2β2=1,\frac{{{x^2}}}{{{\alpha ^2}}} + \frac{{{y^2}}}{{{\beta ^2}}} = 1, the locus of its pole is
View answer and explanation
Correct option: A
x2α2−y2(4a2α2/β2)=1\frac{{{x^2}}}{{{\alpha ^2}}} - \frac{{{y^2}}}{{\left( {4{a^2}{\alpha ^2}/{\beta ^2}} \right)}} = 1
Option A (x2α2−y2(4a2α2/β2)=1\frac{{{x^2}}}{{{\alpha ^2}}} - \frac{{{y^2}}}{{\left( {4{a^2}{\alpha ^2}/{\beta ^2}} \right)}} = 1) is correct because it represents the locus of the pole when the polar with respect to the parabola touches the given ellipse.
409 Medium
409 of 594
Analyzing +4 | -1
If the normals at the points A,B,CA, B, C and DD of an ellipse meet in a point OO, then SA⋅SB⋅SC⋅SDS A \cdot S B \cdot S C \cdot S D is equal to (where, SS is one of the foci)
View answer and explanation
Correct option: A
b2e2⋅SO2\frac{b^2}{e^2} \cdot S O^2
Option A (b2e2⋅SO2\frac{b^2}{e^2} \cdot S O^2) is correct because after substituting and simplifying the expression for the product of distances from the foci to the points, it equals b2e2⋅SO2\frac{b^2}{e^2} \cdot S O^2.
410 Easy
410 of 594
Application +4 | -1
The eccentricity of the conic 4x2+16y2−24x−32y=14{x^2} + 16{y^2} - 24x - 32y = 1 is
View answer and explanation
Correct option: C
32\frac{{\sqrt 3 }}{2}
The correct answer is C (32\frac{{\sqrt 3 }}{2}) because it represents the eccentricity of the given conic section after rewriting it in standard form and identifying the major and minor axes.
411 Easy
411 of 594
Application +4 | -1
The equation of normal at the point (0,3)\left( {0,3} \right) of the ellipse 9x2+5y2=45,9{x^2} + 5{y^2} = 45, is
View answer and explanation
Correct option: B
y−axisy - axis
Option B (y−axisy - axis) is correct because the normal at the point (0,3) of the ellipse is the y-axis.
412 Hard
412 of 594
Analyzing +4 | -1
The locus of the mid-points of chords of the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 that touch the circle x2+y2=b2,{x^2} + {y^2} = {b^2}, is
View answer and explanation
Correct option: B
(x2a2+y2b2)2=b2(x2a4+y2b4){\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}} \right)^2} = {b^2}\left( {\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}}} \right)
Option B (x2a2+y2b2)2=b2(x2a4+y2b4){\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}} \right)^2} = {b^2}\left( {\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}}} \right) is correct because it represents the locus of the midpoints of chords of the ellipse that touch the given circle.
413 Medium
413 of 594
Analyzing +4 | -1
If lx+my+n=0lx + my + n = 0 represents a chord of the ellipse b2x2+a2y2=a2b2{b^2}{x^2} + {a^2}{y^2} = {a^2}{b^2} whose eccentric angles differ by 900{90^0}, then
View answer and explanation
Correct option: C
a2l2+b2m2=2n2{a^2}{l^2} + {b^2}{m^2} = 2{n^2}
Option A (Focus) is correct because the normal chord at (x1, x1) subtends a right angle at the focus of the parabola.
414 Easy
414 of 594
Application +4 | -1
The locus of the middle points of chords of an ellipse which pass through a fixed point is
View answer and explanation
Correct option: A
An ellipse
Option A (An ellipse) is correct because the locus of the midpoints of chords passing through a fixed point in an ellipse is another ellipse.
415 Easy
415 of 594
Application +4 | -1
In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at (0,53),(0,5\sqrt 3 ), then the length of its latus rectum is
View answer and explanation
Correct option: A
5
Option A (5) is correct because, using the given information about the ellipse and its foci, we can solve for the parameters and find the length of the latus rectum.
416 Easy
416 of 594
Analyzing +4 | -1
The tangent at a point P(θ)P(\theta) to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 cuts the auxiliary circle at points QQ and RR. If QRQ R subtends a right angle at the centre CC of the ellipse, then the eccentricity of the ellipse, is
View answer and explanation
Correct option: B
11+sin⁡2θ\frac{1}{\sqrt{1+\sin ^2 \theta}}
Option C (An ellipse) is correct because the locus of the point of intersection of tangents at P and Q to the ellipse is another ellipse.
417 Medium
417 of 594
Application +4 | -1
The eccentricity of the conic 4x2+16y2−24x−32y=14 x^2+16 y^2-24 x-32 y=1 is
View answer and explanation
Correct option: A
32\frac{\sqrt{3}}{2}
Option A (32\frac{{\sqrt{3}}}{{2}}) is correct because, after rewriting the equation in standard form, the eccentricity is calculated using the formula e=1−b2a2e = \sqrt{{1 - \frac{{b^2}}{{a^2}}}}.
418 Easy
418 of 594
Application +4 | -1
If equation of the ellipse is 2x2+3y2−8x+6y+5=02 x^2+3 y^2-8 x+6 y+5=0, then which of the following is true?
View answer and explanation
Correct option: C
Equation of auxiliary circle is x2+y2−4x+2y+2=0x^2+y^2-4 x+2 y+2=0
Option C (Equation of auxiliary circle is x2+y2−4x+2y+2=0x^2+y^2-4 x+2 y+2=0) is correct because after completing the squares and transforming the ellipse equation, the auxiliary circle's equation can be derived.
419 Medium
419 of 594
Analyzing +4 | -1
An ellipse whose focii are (2,4)(2,4) and (14,9)(14,9) and touches x-axis, then its eccentricity is
View answer and explanation
Correct option: C
13313\frac{13}{\sqrt{313}}
Option C (13313\frac{13}{\sqrt{313}}) is correct because it is derived from the geometric properties of the ellipse and the given foci and tangency condition.
420 Easy
420 of 594
Application +4 | -1
If CC is the centre of the ellipse 9x2+16y2=1449{x^2} + 16{y^2} = 144 and SS is one focus. The ratio of CSCS to major axis, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the ratio of CS to the major axis is calculated as 7:8\sqrt{7}:8, which is not among the provided options.
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