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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Easy

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Showing 361-390 of 392 questions

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361 Easy
361 of 392
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.
362 Easy
362 of 392
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.
363 Easy
363 of 392
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}.
364 Easy
364 of 392
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.
365 Easy
365 of 392
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).
366 Easy
366 of 392
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.
367 Easy
367 of 392
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.
368 Easy
368 of 392
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.
369 Easy
369 of 392
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.
370 Easy
370 of 392
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.
371 Easy
371 of 392
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.
372 Easy
372 of 392
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.
373 Easy
373 of 392
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.
374 Easy
374 of 392
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.
375 Easy
375 of 392
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.
376 Easy
376 of 392
Application +4 | -1
Let x=4x = 4 be a directrix to an ellipse whose centre is at origin and its eccentricity is 12 .\frac{1}{2}\,. If P(1, β),   β>0 P\left( {1,\,\beta } \right),\,\,\,\beta > 0\, is a point on this ellipse, then the equation of the normal to it at PP is
View answer and explanation
Correct option: D
4x−2y=14x - 2y = 1
The correct answer is D (4x−2y=14x - 2y = 1) because it is the equation of the normal to the ellipse at point P, derived using the ellipse's properties and the given directrix and eccentricity.
377 Easy
377 of 392
Analyzing +4 | -1
Given ff is increasing, the equation x2f(2a)+y2f(a2−3)=1\frac{x^2}{f(2 a)}+\frac{y^2}{f\left(a^2-3\right)}=1 represents an ellipse with XX-axis as major axis if
View answer and explanation
Correct option: C
(−1,3)(-1,3)
Option C (−1,3)(-1,3) is correct because it represents the interval where aa satisfies the inequality derived from the condition for the ellipse to have the X-axis as the major axis.
378 Easy
378 of 392
Application +4 | -1
A point on the ellipse x216+y29=1\frac{x^2}{16}+\frac{y^2}{9}=1 at a distance equal to the mean of lengths of the semi-major and semi-minor axis from the centre, is
View answer and explanation
Correct option: A
(2917,310514)\left(\frac{2 \sqrt{91}}{7}, \frac{3 \sqrt{105}}{14}\right)
Option A ((2917,310514)\left(\frac{2 \sqrt{91}}{7}, \frac{3 \sqrt{105}}{14}\right)) is correct because substituting the parametric coordinates into the distance equation and solving yields this point.
379 Easy
379 of 392
Remembering +4 | -1
Length of latusrectum 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
2a2b\frac{2 a^2}{b}, if a<ba < b
Option C (2b2a\frac{2 b^2}{a}, if a<ba < b) is incorrect. Option A (2a2b\frac{2 a^2}{b}, if a<ba < b) is correct because the length of the latus rectum of an ellipse when a < b is 2a2b\frac{2a^2}{b}.
380 Easy
380 of 392
Understanding +4 | -1
The curve with parametric equations x=α+5cos⁡θ,  y=β+4sin⁡θx = \mathrm{\alpha } + 5\cos \mathrm{\theta },\;y = \mathrm{\beta } + 4\sin \mathrm{\theta } (where θ\theta is parameter) is
View answer and explanation
Correct option: B
An ellipse
Option B (An ellipse) is correct because the parametric equations describe an ellipse centered at (α, β) with semi-major and semi-minor axes of 5 and 4, respectively.
381 Easy
381 of 392
Application +4 | -1
The eccentricity of the hyperbola whose length of the latusrectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is
View answer and explanation
Correct option: C
23\frac{2}{{\sqrt 3 }}
Option C (23\frac{2}{{\sqrt 3 }}) is correct because it is the eccentricity calculated from the given conditions on the latus rectum and conjugate axis.
382 Easy
382 of 392
Application +4 | -1
The eccentric angle of a point on the ellipse x26+y22=1\frac{{{x^2}}}{6} + \frac{{{y^2}}}{2} = 1, whose distance from the centre of the ellipse is 2, is
View answer and explanation
Correct option: A
π4\frac{\pi }{4}
The correct answer is A (π4\frac{\pi}{4}) because substituting the parametric form of the ellipse into the distance formula and solving for the eccentric angle yields this result.
383 Easy
383 of 392
Application +4 | -1
If the eccentricity of the hyperbola x2−y2coces⁡2α=25x^2-y^2 \operatorname{coces}^2 \alpha=25 is 5\sqrt{5} times the eccentricity of the ellipse x2cosec⁡2α+y2=5x^2 \operatorname{cosec}^2 \alpha+y^2=5, then α\alpha is equal to :
View answer and explanation
Correct option: A
tan⁡−12\tan ^{-1} \sqrt{2}
Option A (tan⁡−12\tan^{-1} \sqrt{2}) is correct because by equating the eccentricities of the hyperbola and the ellipse as given in the question, we can solve for α\alpha and find that α=tan⁡−12\alpha = \tan^{-1} \sqrt{2}.
384 Easy
384 of 392
Application +4 | -1
The eccentricity of an ellipse x2a2+y2b2=1\frac{x^{2}}{a^2}+\frac{y^2}{b^2}=1 whose latusrectum is half of its minor axis is
View answer and explanation
Correct option: C
32\frac{\sqrt{3}}{2}
Option C (32\frac{{\sqrt{{3}}}}{{2}}) is correct because the given condition (latus rectum is half of its minor axis) leads to the relationship a=2ba = 2b, which allows us to calculate the eccentricity as 1−b2a2=1−14=32\sqrt{{1 - \frac{{b^2}}{{a^2}}}} = \sqrt{{1 - \frac{{1}}{{4}}}} = \frac{{\sqrt{{3}}}}{{2}}.
385 Easy
385 of 392
Application +4 | -1
For an ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=1 with vertices AA and A′A^{\prime}, tangent drawn at the point PP in the first quadrant meets the yy-axis in Q\mathrm{Q} and the chord A′P\mathrm{A}^{\prime} \mathrm{P} meets the y\mathrm{y}-axis in M\mathrm{M}. If ' O\mathrm{O} ' is the origin then OQ2−MQ2\mathrm{OQ}^2-\mathrm{MQ}^2 equals to
View answer and explanation
Correct option: C
4
Option B (by = ax - √(a4+a2 b2+b4)) is correct because it satisfies the condition for a common tangent to both ellipses.
386 Easy
386 of 392
Application +4 | -1
The eccentricity of the ellipse 9x2+5y2−18x−20y−16=09{x^2} + 5{y^2} - 18x - 20y - 16 = 0 is
View answer and explanation
Correct option: B
23\frac{2}{3}
Option B (23\frac{2}{3}) is correct because it's the eccentricity calculated after rewriting the ellipse equation in standard form and applying the eccentricity formula.
387 Easy
387 of 392
Application +4 | -1
If the point pp on the ellipse 4x2+5y2=204 x^2+5 y^2=20 is farthest from the point Q(0,−4)Q(0,-4), then PQ2P Q^2 is equal to
View answer and explanation
Correct option: D
36
Option D (36) is correct because it represents the maximum squared distance between a point on the ellipse and the given point Q(0, -4).
388 Easy
388 of 392
Application +4 | -1
The length of the chord of the ellipse x225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1, where mid-point is (12,23)\left(\frac{1}{2}, \frac{2}{3}\right) is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated length of the chord does not match any of the provided options.
389 Easy
389 of 392
Application +4 | -1
The eccentricity of an ellipse, with centre at the origin is 2/3.2/3. If one of directrices is x=6,x = 6, then equation of the ellipse is
View answer and explanation
Correct option: C
5x2+9y2=805x^2 + 9y^2 = 80
Option C (5x2+9y2=805x^2 + 9y^2 = 80) is correct because it is derived by using the given eccentricity and directrix to find the equation of the ellipse.
390 Easy
390 of 392
Analyzing +4 | -1
The curve represented by the equation 4x2+16y2−24x−32y−12=04{x^2} + 16{y^2} - 24x - 32y - 12 = 0 is
View answer and explanation
Correct option: D
An ellipse with eccentricity 3/2\sqrt 3 /2
Option D (An ellipse with eccentricity 3/2\sqrt 3 /2) is correct because the equation represents an ellipse, and its eccentricity is calculated as 1−(b2/a2)\sqrt{1 - (b^2/a^2)}.
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