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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 421-450 of 594 questions

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421 Medium
421 of 594
Application +4 | -1
Coordinates of the foci of the ellipse 2x2+3y2−4x−12y+13=02 x^2+3 y^2-4 x-12 y+13=0 are
View answer and explanation
Correct option: A
(1±16,2)\left(1 \pm \frac{1}{\sqrt{6}}, 2\right)
Option A ((1±16,2)\left(1 \pm \frac{{1}}{{\sqrt{{6}}}}, 2\right)) is correct because it is derived by completing the square, finding the ellipse parameters, and calculating the foci coordinates.
422 Easy
422 of 594
Understanding +4 | -1
PP is any point on the ellipse 81x2+144y2=194481{x^2} + 144{y^2} = 1944, whose foci are SS and S′S'. Then, SP+S′PSP + S'P equals
View answer and explanation
Correct option: B
464\sqrt 6
Option B (464\sqrt 6) is correct because SP+S′PSP + S'P equals to 2a2a which is 464\sqrt{6} based on the given ellipse equation.
423 Medium
423 of 594
Analyzing +4 | -1
If the normal at any point PP of the ellipse x2a2+y2b2=1(a>b>0)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0) meets the major and minor axes at G{G} and G′G^{\prime} respectively and OF{OF} is the perpendicular drawn from centre O{O} to this normal, then PF⋅PGPF \cdot PG is equal to
View answer and explanation
Correct option: A
b2{b}^2
The correct answer is A (b2{b}^2). By finding the coordinates of G and using the perpendicular distance formula, we can derive that PF⋅PG=b2PF \cdot PG = b^2.
424 Easy
424 of 594
Analyzing +4 | -1
If the ellipse x24+y2=1\frac{x^2}{4}+y^2=1 meets the ellipse x2+y2a2=1x^2+\frac{y^2}{a^2}=1 in four distinct points and a=b2−5b+7a=b^2-5 b+7, then bb belongs to
View answer and explanation
Correct option: B
(−∞,2)∪(3,∞)(-\infty, 2) \cup(3, \infty)
Option C ((2,3)(2,3)) is correct because the condition for four distinct intersection points implies that a>1a > 1, leading to the inequality b2−5b+6>0b^2 - 5b + 6 > 0, which gives the solution b in (2,3).
425 Easy
425 of 594
Application +4 | -1
The point (4,−3)(4,-3) w.r.t. the ellipse 4x2+5y2=14 x^2+5 y^2=1
View answer and explanation
Correct option: C
Is outside the curve
Option C (Is outside the curve) is correct because substituting the point into the ellipse equation results in a value greater than 1, indicating it lies outside.
426 Easy
426 of 594
Application +4 | -1
For the ellipse 3x2+4y2=123 x^2+4 y^2=12, the length of latus rectum is
View answer and explanation
Correct option: B
3
Option B (3) is correct because after converting the ellipse equation to standard form, the latus rectum is calculated as 2b2a\frac{{2b^2}}{{a}}.
427 Easy
427 of 594
Understanding +4 | -1
The equation 2x2+3y2=302 x^2+3 y^2=30 represents
View answer and explanation
Correct option: B
An ellipse
Option B (An ellipse) is correct because the equation can be rewritten in the standard form of an ellipse equation.
428 Easy
428 of 594
Application +4 | -1
If SS and S′S^{\prime} are two foci of the ellipse 16x2+25y2=40016 x^2+25 y^2=400 and PSQ is a focal chord such that SP=16S P=16, then SQ=S Q=
View answer and explanation
Correct option: D
74/974 / 9
Option D (74/974 / 9) is correct because it is derived using the harmonic mean of focal segments and the property that the sum of distances from a point on the ellipse to the foci equals the major axis.
429 Easy
429 of 594
Application +4 | -1
Area of quadrilateral formed by joining the focii of the ellipse, x216+y212=1\frac{x^2}{16}+\frac{y^2}{12}=1 and x212+y216=1\frac{x^2}{12}+\frac{y^2}{16}=1 is (in sq. units)
View answer and explanation
Correct option: D
8
Option D (2xyℓn(xy)=x2+C2 x y \ell n(x y)=x^2+C) is correct because by rearranging the differential equation and integrating, we arrive at the given solution.
430 Medium
430 of 594
Application +4 | -1
If the tangents are drawn to the ellipse x2+2y2=2{x^2} + 2{y^2} = 2, then the locus of the mid point of the intercept made by the tangents between the coordinate axes is
View answer and explanation
Correct option: A
12x2+14y2=1\frac{1}{{2{x^2}}} + \frac{1}{{4{y^2}}} = 1
Option A (12x2+14y2=1\frac{1}{{2{x^2}}} + \frac{1}{{4{y^2}}} = 1) is correct because it represents the locus of the midpoint of the intercept made by the tangents between the coordinate axes, derived using the equation of the tangent to the ellipse.
431 Medium
431 of 594
Analyzing +4 | -1
If the line xcos⁡α+ysin⁡α=px\cos \mathrm{\alpha } + y\sin \mathrm{\alpha } = p be normal to the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1, then
View answer and explanation
Correct option: D
p2(a2sec⁡2α+b2cosec2α)=(a2−b2)2{p^2}\left( {{a^2}{{\sec }^2}\mathrm{\alpha } + {b^2}\mathrm{cose}{\mathrm{c}^2}\mathrm{\alpha }} \right) = {\left( {{a^2} - {b^2}} \right)^2}
Option D (p2(a2sec⁡2α+b2cosec2α)=(a2−b2)2{p^2}\left( {{a^2}{{\sec }^2}\alpha + {b^2}\mathrm{cose}{\mathrm{c}^2}\alpha } \right) = {\left( {{a^2} - {b^2}} \right)^2}) is correct because it is derived by equating the normal equation of the ellipse with the given line equation.
432 Easy
432 of 594
Application +4 | -1
If y=xy=x and 3y+2x=03 y+2 x=0 are the equations of a pair of conjugate diameters of an ellipse, then the eccentricity of the ellipse is
View answer and explanation
Correct option: B
1/31 / \sqrt{3}
Option B (1/31 / \sqrt{3}) is correct because it's derived from the relationship between the slopes of conjugate diameters and the ellipse's eccentricity.
433 Easy
433 of 594
Application +4 | -1
In triangle ABC\mathrm{ABC} given the base BC\mathrm{BC} of the triangle and if ∠C−∠B=K\angle C-\angle{B}=K, a constant. Then the locus of the vertex A\mathrm{A} is
View answer and explanation
Correct option: A
hyperbola
Option A (hyperbola) is correct because the locus of the vertex A, given the base BC and a constant difference between angles C and B, results in the equation of a hyperbola.
434 Easy
434 of 594
Application +4 | -1
Equation of the ellipse whose axes are the axes of co-ordinates and which passes through the point (−3,1)(-3,1) and has eccentricity 25\sqrt{\frac{2}{5}}, is
View answer and explanation
Correct option: D
5x2+3y2−32=05 x^2+3 y^2-32=0
Option D (5x2+3y2−32=05 x^2+3 y^2-32=0) is correct because it is the equation of the ellipse that passes through the given point and has the given eccentricity.
435 Medium
435 of 594
Analyzing +4 | -1
Let the ellipse E:x2+9y2=9E: x^2+9 y^2=9 intersect the positive xx - and yy-axes at the points AA and BB respectively. Let the major axis of EE be a diameter of the circle CC. Let the line passing through AA and BB meet the circle CC at the point PP. If the area of the triangle with vertices A,PA, P and the origin OO is mn\frac{m}{n}, where mm and nn are coprime, then m−nm-n is equal to
View answer and explanation
Correct option: C
17
The correct answer is C (17) because the area of the triangle AOP is calculated using the intersection of the ellipse, circle, and line equations, resulting in an area of 27/10, and thus m-n = 27-10 = 17.
436 Easy
436 of 594
Application +4 | -1
The locus of the foot of the perpendicular drawn from the centre of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 on any tangent, is
View answer and explanation
Correct option: C
(x2+y2)2=a2x2+b2y2\left(x^2+y^2\right)^2=a^2 x^2+b^2 y^2
Option C ((x2+y2)2=a2x2+b2y2(x^2+y^2)^2=a^2 x^2+b^2 y^2) is correct because it represents the locus of the foot of the perpendicular from the center of the ellipse on any tangent.
437 Medium
437 of 594
Application +4 | -1
If mm is the slope of a common tangent to the curves x216+y29=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1 and  x2+y2=12,\,{{\text{x}}^2} + {y^2} = 12, then 12m2{\text{12}}{m^2} is equal to:
View answer and explanation
Correct option: B
9
Option B (9) is correct because after finding the condition for a common tangent and solving for m2m^2, we find that 12m2=912m^2 = 9.
438 Easy
438 of 594
Application +4 | -1
If the foci of the ellipse x216+y2b2=1\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1 coincide with the foci of the hyperbola x2144−y281=125\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}, then b2b^{2} is equal to
View answer and explanation
Correct option: C
7
Option C (7) is correct because the value of b2b^2 is calculated by equating the foci of the ellipse and hyperbola.
439 Medium
439 of 594
Application +4 | -1
The distance of the point 'θ\theta' on the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 from a focus is
View answer and explanation
Correct option: C
a(1+ecos⁡θ)a\left( {1 + e\cos \theta } \right)
Option C, a(1+ecos⁡θ)a(1 + e\cos \theta), is correct because the focal distance is derived using the ellipse equation and the definition of eccentricity.
440 Easy
440 of 594
Application +4 | -1
The line 12xcos⁡θ + 5ysin⁡θ = 6012x\cos \theta \, + \,5y\sin \theta \, = \,60 is tangent to which of the following curves?
View answer and explanation
Correct option: B
144x2+25y2=3600144{x^2} + 25{y^2} = 3600
Option B (144x2+25y2=3600144{x^2} + 25{y^2} = 3600) is correct because the given line is tangent to the ellipse x225+y2144=1\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{144}} = 1, which is equivalent to the given equation.
441 Medium
441 of 594
Application +4 | -1
The locus of the foot of the perpendicular from the foci on any tangent to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, is
View answer and explanation
Correct option: B
x2+y2=a2x^2+y^2=a^2
Option B (x2+y2=a2x^2+y^2=a^2) is correct because it represents the locus of the foot of the perpendicular from the foci on any tangent to the ellipse, which is the auxiliary circle.
442 Medium
442 of 594
Application +4 | -1
The equation of the normal to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 at the positive end of the latusrectum, is
View answer and explanation
Correct option: B
x−ey−e3a=0x-e y-e^3 a=0
Option B (x−ey−e3a=0x-e y-e^3 a=0) is correct because it represents the equation of the normal to the ellipse at the positive end of the latus rectum after substituting the coordinates and simplifying.
443 Easy
443 of 594
Understanding +4 | -1
If AA and BB are two fixed points and PP is a variable point such that PA+PB=4,PA + PB = 4, the locus of PP is
View answer and explanation
Correct option: B
An ellipse
Option B (An ellipse) is correct because the sum of distances from two fixed points being constant defines an ellipse.
444 Easy
444 of 594
Application +4 | -1
The eccentricity of the ellipse 9x2+5y2−30y=09{x^2} + 5{y^2} - 30y = 0 is
View answer and explanation
Correct option: B
23\frac{2}{3}
Option B (23\frac{2}{3}) is correct because the eccentricity of the ellipse is calculated using the formula e=1−a2b2e = \sqrt{1 - \frac{a^2}{b^2}} after rewriting the given equation in standard form.
445 Easy
445 of 594
Application +4 | -1
The line y=x+1y = x + 1 meets the ellipse x24+y22=1\frac{{{x^2}}}{4} + \frac{{{y^2}}}{2} = 1 at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)2{(3r)^2} is equal to:
View answer and explanation
Correct option: A
20
Option B (9) is correct because it is the value of 12m2, where m is the slope of the common tangent to the given ellipse and circle, calculated by equating the tangent equations.
446 Medium
446 of 594
Application +4 | -1
The equation of the tangents which are drawn at the ends of the major axis of the ellipse 9x2+5y2−30y=09 x^2+5 y^2-30 y=0 are
View answer and explanation
Correct option: C
y=0,y=6y=0, y=6
Option C (y=0,y=6y=0, y=6) is correct because these lines represent the tangents at the ends of the major axis of the given ellipse.
447 Medium
447 of 594
Analyzing +4 | -1
If y=mx−(a2−b2)ma2+b2m2y = mx - \frac{{\left( {{a^2} - {b^2}} \right)m}}{{\sqrt {{a^2} + {b^2}{m^2}} }} is normal to the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 for all values of mm belonging to
View answer and explanation
Correct option: C
R
Option C (R) is correct because the condition for the line to be normal to the ellipse holds for all real values of m, which is the slope of the normal.
448 Easy
448 of 594
Application +4 | -1
If α\alpha and β\beta are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse, is
View answer and explanation
Correct option: D
sin⁡α+sin⁡βsin⁡(α+β)\frac{\sin \alpha+\sin \beta}{\sin (\alpha+\beta)}
Option D (sin⁡α+sin⁡βsin⁡(α+β)\frac{{\sin \alpha + \sin \beta}}{{\sin (\alpha + \beta)}}) is correct because the eccentricity is derived using the equation of a focal chord and the properties of eccentric angles.
449 Easy
449 of 594
Understanding +4 | -1
The sum of the focal distances from any point on the ellipse 9x2+16y2=1449{x^2} + 16{y^2} = 144 is
View answer and explanation
Correct option: C
8
Option C (8) is correct because the sum of the focal distances from any point on the ellipse is equal to the length of its major axis (2a).
450 Hard
450 of 594
Application +4 | -1
If the focal distance of an end of the minor axis of an ellipse (referred to its axes as the axes of xx and yy respectively) is kk and the distance between its foci is 2h2h, then its equation is
View answer and explanation
Correct option: B
x2k2+y2k2−h2=1\frac{{{x^2}}}{{{k^2}}} + \frac{{{y^2}}}{{{k^2} - {h^2}}} = 1
Option B (x2k2+y2k2−h2=1\frac{{{x^2}}}{{{k^2}}} + \frac{{{y^2}}}{{{k^2} - {h^2}}} = 1) is correct because it correctly relates the focal distance and the distance between the foci to the equation of the ellipse.
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