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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Medium

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Showing 1-30 of 187 questions
1 Medium
1 of 187
Application +4 | -1
Let x2144+y281=1\frac{x^2}{144}+\frac{y^2}{81}=1 be the equation of an ellipse. If mm be the minimum value of the segment of tangent to the ellipse intercepted by the coordinate axes and M\mathrm{M} be the maximum distance of the centre of the ellipse from the normal, then the value of mM\mathrm{mM} is equal to
View answer and explanation
Correct option: A
63
Option A (63) is correct because it involves finding the minimum segment of the tangent intercepted by the coordinate axes and the maximum distance of the center from the normal, then multiplying these values.
2 Medium
2 of 187
Application +4 | -1
If the co-ordinates of two points AA and BB are (7,0)(\sqrt{7}, 0) and (−7,0)(-\sqrt{7}, 0) respectively and PP is any point on the conic, 9x2+16y2=1449 x^{2}+16 y^{2}=144, then PA+PB\mathrm{PA}+\mathrm{PB} is equal to :
View answer and explanation
Correct option: B
8
Option B (8) is correct because PA+PB=2aPA + PB = 2a for any point P on the ellipse, where A and B are the foci.
3 Medium
3 of 187
Analyzing +4 | -1
Any ordinate NPN P of an ellipse meets the auxiliary circle at QQ. Then, the locus of the intersection of the normal at PP and OO is the circle
View answer and explanation
Correct option: B
x2+y2=(a+b)2x^2+y^2=(a+b)^2
Option B (x2+y2=(a+b)2x^2+y^2=(a+b)^2) is correct because it represents the locus of the intersection of the normal at P and Q, derived using trigonometric parameterization and simplification.
4 Medium
4 of 187
Application +4 | -1
The perpendicular from the centre of an ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 to polar of a point with respect to the ellipse is constant equal to cc. Then, the locus of the point is the ellipse
View answer and explanation
Correct option: C
x2a4+y2b4=1c2\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{c^2}
Option C (x2a4+y2b4=1c2\frac{{x^2}}{{a^4}} + \frac{{y^2}}{{b^4}} = \frac{{1}}{{c^2}}) is correct because it represents the locus of the point when the perpendicular distance from the center to the polar is constant.
5 Medium
5 of 187
Application +4 | -1
The locus of the pole of any tangent to the auxiliary circle w.r.t. the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is
View answer and explanation
Correct option: D
b4x2+a4y2=a2b4b^4 x^2+a^4 y^2=a^2 b^4
Option A (b4x2+a4y2=a2b4b^4 x^2+a^4 y^2=a^2 b^4) is correct because it is derived by finding the pole of a tangent to the auxiliary circle with respect to the ellipse.
6 Medium
6 of 187
Analyzing +4 | -1
The locus of the poles of tangents to the director circle of the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 with respect 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
x2a4+y2a4=1a2+b2\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{a^4}}} = \frac{1}{{{a^2} + {b^2}}}
Option B (x2a4+y2b4=1a2+b2\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} = \frac{1}{{{a^2} + {b^2}}}) is correct because it represents the locus of the poles of tangents to the director circle with respect to the given ellipse.
7 Medium
7 of 187
Application +4 | -1
One of the directrices of the ellipse 8x2+6y2−16x+12y+13=08{x^2} + 6{y^2} - 16x + 12y + 13 = 0 is
View answer and explanation
Correct option: B
3y+3=63y + 3 = \sqrt 6
Option B (3y+3=63y + 3 = \sqrt 6) is correct because it represents one of the directrices of the given ellipse after completing the square and applying the directrix formula.
8 Medium
8 of 187
Analyzing +4 | -1
If the chords of contact of tangents from two points (x1,y1)\left(x_1, y_1\right) and (x2,y2)\left(x_2, y_2\right) to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 are at right angles, then x1x2y1y2\frac{x_1 x_2}{y_1 y_2} is equal to
View answer and explanation
Correct option: C
−a4b4-\frac{a^4}{b^4}
Option D (None of these) is correct because the calculated distance between the tangents does not fall within the given ranges.
9 Medium
9 of 187
Application +4 | -1
Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=1 from any of its foci
View answer and explanation
Correct option: C
(1,22)(1,2 \sqrt{2})
Option C ((1,22)(1,2 \sqrt{2})) is correct because the locus of the foot of the perpendicular from the foci to any tangent of the ellipse is the auxiliary circle, x2+y2=a2x^2 + y^2 = a^2, where a=3. Substituting the point (1,22)(1,2\sqrt{2}) satisfies the equation x2+y2=9x^2 + y^2 = 9.
10 Medium
10 of 187
Analyzing +4 | -1
If circumcentre of an equilateral triangle inscribed in x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, with vertices having eccentric angle α,β,γ\alpha, \beta, \gamma respectively is (h,k)(h, k), then Σcos⁡(α−β)=\Sigma \cos (\alpha-\beta)=
View answer and explanation
Correct option: D
9h22a2+9k22b2−32\frac{9 h^2}{2 a^2}+\frac{9 k^2}{2 b^2}-\frac{3}{2}
Option D (9h22a2+9k22b2−32\frac{{9 h^2}}{{2 a^2}}+\frac{{9 k^2}}{{2 b^2}}-\frac{{3}}{{2}}) is correct because it relates the circumcenter coordinates (h, k) to the eccentric angles and the ellipse parameters.
11 Medium
11 of 187
Application +4 | -1
If BB is an end of the minor axis of the ellipse x2a2+y2b2=1,F1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, F_1 and F2F_2 are the foci such that the triangle BF1F2BF_1F_2 is equilateral, then eccentricity of the ellipse is:
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{{1}}{{2}}) is correct because the eccentricity is derived from the equilateral triangle condition.
12 Medium
12 of 187
Application +4 | -1
Let P\mathrm{P} be a variable point on the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 with foci F1F_1 and F2F_2. If A\mathrm{A} is the area of triangle PF1F2P F_1 F_2 then the maximum value of AA is
View answer and explanation
Correct option: B
aeb
Option B (aeb) is correct because the maximum area of the triangle is obtained when sin(theta) = 1, resulting in abe.
13 Medium
13 of 187
Application +4 | -1
On the ellipse 4x2+9y2=14 x^2+9 y^2=1, the points at which the tangents are parallel to the line 8x=9y8 x=9 y, are
View answer and explanation
Correct option: B
(±2/5,±1/5)(\pm 2 / 5, \pm 1 / 5)
Option D ((±25,∓15)\left(\pm \frac{2}{5}, \mp \frac{1}{5}\right)) is correct because using the formula for points on the ellipse where tangents have a given slope, we find the points to be (±25,∓15)\left(\pm \frac{2}{5}, \mp \frac{1}{5}\right).
14 Medium
14 of 187
Application +4 | -1
The equation of the ellipse, whose vertices are (±5,0)( \pm 5,0) and foci are (±4,0)( \pm 4,0), is
View answer and explanation
Correct option: A
9x2+25y2=2259 x^2+25 y^2=225
Option B (4x281+4y245=1\frac{4 x^2}{81}+\frac{4 y^2}{45}=1) is correct because it is found using the given eccentricity and latus rectum.
15 Medium
15 of 187
Application +4 | -1
The locus of point of intersection of perpendicular tangents of ellipse (x−1)216+(y−1)29=1\frac{(x-1)^2}{16}+\frac{(y-1)^2}{9}=1 is -
View answer and explanation
Correct option: C
x2+y2−2x−2y−23=0x^2+y^2-2 x-2 y-23=0
Option C (x2+y2−2x−2y−23=0x^2+y^2-2 x-2 y-23=0) is correct because the locus of perpendicular tangents of an ellipse is its director circle, which is found by (x−h)2+(y−k)2=a2+b2(x-h)^2 + (y-k)^2 = a^2 + b^2.

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16 Medium
16 of 187
Application +4 | -1
In an ellipse, the distance between the foci is 6 and length of semi minor axis is 4, then eccentricity of the ellipse is
View answer and explanation
Correct option: B
35\frac{3}{5}
Option B (35\frac{3}{5}) is correct because using the given information about the distance between foci and the semi-minor axis, we can find the eccentricity using the relationship between a, b, and e.
17 Medium
17 of 187
Application +4 | -1
If the normal at the P(θ)P\left( \theta \right) to the ellipse x214+y25=1\frac{{{x^2}}}{{14}} + \frac{{{y^2}}}{5} = 1 intersect it again at the point Q(2θ)Q\left( {2\mathrm{\theta }} \right), then cos⁡θ\cos \theta is equal to
View answer and explanation
Correct option: B
−23- \frac{2}{3}
Option B (−23\frac{{ - 2}}{3}) is correct because it satisfies the condition derived from the normal equation intersecting the ellipse at two points.
18 Medium
18 of 187
Analyzing +4 | -1
A triangle is drawn such that it is right angled at the centre of the ellipse x2a2+y2b2=1(a>b)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a > b) and its other two vertices lie on the ellipse with eccentric angles α,β\alpha, \beta. Then, 1−e2cos⁡2(α+β2)cos⁡2(α+β2)\frac{1-e^2 \cos ^2\left(\frac{\alpha+\beta}{2}\right)}{\cos ^2\left(\frac{\alpha+\beta}{2}\right)} is equal to
View answer and explanation
Correct option: B
a2+b2a2\frac{a^2+b^2}{a^2}
Option B (a2+b2a2\frac{a^2+b^2}{a^2}) is correct because it represents the given expression after homogenizing the equation of the ellipse and using the condition for perpendicular lines.
19 Medium
19 of 187
Application +4 | -1
If the tangent at any point PP on the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 meets the tangents at the vertices AA and A′A' in LL and L′L' respectively, then AL.A′L′=AL.A'L' =
View answer and explanation
Correct option: D
b2{b^2}
Option D (b2{b^2}) is correct because the product of the lengths AL and A'L' is equal to b2, a property of the ellipse.
20 Medium
20 of 187
Application +4 | -1
If y=x+cy=x+c is a normal to the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=1, then c2c^2 is equal to
View answer and explanation
Correct option: B
2513\frac{25}{13}
Option B (2513\frac{25}{13}) is correct because it is the value of c2c^2 when y=x+cy=x+c is a normal to the given ellipse.
21 Medium
21 of 187
Analyzing +4 | -1
The minimum \& maximum distances of a point (−1,6)(-1,6) from the ellipse 25x2+9y2+50x+36y−164=025 x^2+9 y^2+50 x+36 y-164=0 are ll and mm respectively then m2−l2m^2-l^2 equals
View answer and explanation
Correct option: D
160
Option D (160) is correct because it calculates the difference of squares of the maximum and minimum distances from the point to the ellipse after rewriting the ellipse equation in standard form.
22 Medium
22 of 187
Understanding +4 | -1
If PP is any point lying on the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, whose foci are SS and S′S^{\prime}. Let ∠PSS′=α\angle P S S^{\prime}=\alpha and ∠PS′S=β\angle P S^{\prime} S=\beta, then which of the following is/are true?
View answer and explanation
Correct option: D
tan⁡α2tan⁡β2=a2−b2b2(a−a2−b2)\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{\sqrt{a^2-b^2}}{b^2}\left(a-\sqrt{a^2-b^2}\right)
Option D (tan⁡α2tan⁡β2=a2−b2b2(a−a2−b2)\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{\sqrt{{a^2-b^2}}}{b^2}\left(a-\sqrt{{a^2-b^2}}\right)) is correct because it is derived using the properties of the ellipse and trigonometric relationships involving the angles to the foci.
23 Medium
23 of 187
Application +4 | -1
3x2+4y2–6x+8y+k=03x^2 + 4y^2 – 6x + 8y + k = 0 represents an ellipse with eccentricity 1/2,1/2,
View answer and explanation
Correct option: C
k<7k < 7
Option C (k<7k < 7) is correct because the given equation represents an ellipse only when k<7k < 7, ensuring the terms inside the square roots are positive.
24 Medium
24 of 187
Application +4 | -1
If the lengths of major and semi-minor axes of an ellipse are 4 and 3\sqrt 3 and their corresponding equations are y−5=0y - 5 = 0 and x+3=0,x + 3 = 0, then the equation of the ellipse is
View answer and explanation
Correct option: A
3x2+4y2+18x−40y+115=03{x^2} + 4{y^2} + 18x - 40y + 115 = 0
Option A (3x2+4y2+18x−40y+115=03{x^2} + 4{y^2} + 18x - 40y + 115 = 0) is correct because it represents the equation of the ellipse with given major and semi-minor axes and center coordinates.
25 Medium
25 of 187
Analyzing +4 | -1
The locus of the intersection of two normal to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 which are perpendicular to each other, is
View answer and explanation
Correct option: A
(a2−b2)2(x2a2−y2b2)2=(a2+b2)2(x2+y2)(x2a2+y2b2)2\left(a^2-b^2\right)^2\left(\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)^2=\left(a^2+b^2\right)^2\left(x^2+y^2\right)\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)^2
Option A is correct because it represents the locus of the intersection of two perpendicular normals to the ellipse, derived through eliminating parameters from normal equations.
26 Medium
26 of 187
Analyzing +4 | -1
A line intersects the ellipse x24a2+y2a2=1\frac{x^2}{4 a^2}+\frac{y^2}{a^2}=1 at AA and BB and the parabola y2=4a(x+2a)y^2=4 a(x+2 a) at CC and DD. The line segment ABA B subtends a right angle at the centre of the ellipse. Then, the locus of the point of intersection of tangents to the parabola at CC and DD, is
View answer and explanation
Correct option: D
y2+4a2=5(x+4a)2y^2+4 a^2=5(x+4 a)^2
The correct answer is D (y2+4a2=5(x+4a)2y^2+4 a^2=5(x+4 a)^2) because it represents the locus of the intersection point of tangents to the parabola, derived using the condition that the chord AB subtends a right angle at the center of the ellipse.
27 Medium
27 of 187
Application +4 | -1
Chord of the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 touch x2α2+y2β2=1,\frac{{{x^2}}}{{{\alpha ^2}}} + \frac{{{y^2}}}{{{\beta ^2}}} = 1, then the locus of their pole is
View answer and explanation
Correct option: C
α2a4x2+β2b4y2=1\frac{\alpha^2}{a^4} x^2+\frac{\beta^2}{b^4} y^2=1
Option C (α2a4x2+β2b4y2=1\frac{\alpha^2}{a^4} x^2+\frac{\beta^2}{b^4} y^2=1) is correct because it represents the locus of the pole of a chord of the given ellipse that touches another ellipse.
28 Medium
28 of 187
Analyzing +4 | -1
The locus of the centroid of an equilateral triangle inscribed in the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is
View answer and explanation
Correct option: D
(a2+3b2)2a2x2+(b2+3a2)2b2y2=(a2−b2)2\frac{\left(a^2+3 b^2\right)^2}{a^2} x^2+\frac{\left(b^2+3 a^2\right)^2}{b^2} y^2=\left(a^2-b^2\right)^2
Option D ((a2+3b2)2a2x2+(b2+3a2)2b2y2=(a2−b2)2\frac{\left(a^2+3 b^2\right)^2}{a^2} x^2+\frac{\left(b^2+3 a^2\right)^2}{b^2} y^2=\left(a^2-b^2\right)^2) is correct because the locus of the centroid of an equilateral triangle inscribed in the ellipse is given by this equation, derived by using the properties of the centroid and the ellipse equation.
29 Medium
29 of 187
Application +4 | -1
Let f(x)=x2+9,g(x)=xx−9f(x)=x^2+9, g(x)=\frac{x}{x-9} and a=fog(10),b=gof(3)a=fog(10), b=g o f(3). If ee and ll denote the eccentricity and the length of the latus rectum of the ellipse x2a+y2b=1\frac{x^2}{a}+\frac{y^2}{b}=1, then 8e2+l28 e^2+l^2 is equal to.
View answer and explanation
Correct option: B
8
Option B (8) is correct because it is the value of the expression 8e2 + l2 after calculating the eccentricity and latus rectum of the ellipse.
30 Medium
30 of 187
Application +4 | -1
The locus of the point of intersection of tangents to an ellipse at two points, sum of whose eccentric angles is constant is a/an
View answer and explanation
Correct option: D
straight line
Option D (straight line) is correct because the locus of the point of intersection of tangents is a straight line when the sum of the eccentric angles is constant.
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