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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Mixed difficulty

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31 Easy
31 of 594
Application +4 | -1
The eccentricity of the ellipse x2+4y2+8y−2x+1=0x^2+4 y^2+8 y-2 x+1=0, is
View answer and explanation
Correct option: A
32\frac{\sqrt{3}}{2}
Option A (32\frac{\sqrt{3}}{2}) is correct because the eccentricity is calculated using the formula e=1−b2a2e = \sqrt{1 - \frac{b^2}{a^2}} after rewriting the ellipse equation in standard form.
32 Easy
32 of 594
Application +4 | -1
The distance of the centre of ellipse x2+2y2−2=0{x^2} + 2{y^2} - 2 = 0 to those tangents of the ellipse which are equally inclined from both the axes, is
View answer and explanation
Correct option: D
32\frac{{\sqrt 3 }}{2}
Option B (32\sqrt {\frac{3}{2}}) is incorrect. Option D (32\frac{{\sqrt 3 }}{2}) is correct because the distance from the center of the ellipse to the tangents equally inclined to both axes is 32\frac{{\sqrt 3 }}{2}.
33 Easy
33 of 594
Application +4 | -1
The centre of the ellipse (x+y−2)29+(x−y)216=1\frac{{{{\left( {x + y - 2} \right)}^2}}}{9} + \frac{{{{\left( {x - y} \right)}^2}}}{{16}} = 1 is
View answer and explanation
Correct option: B
(1,1)\left( {1,1} \right)
The correct answer is B ((1,1)) because it represents the center of the given ellipse.
34 Easy
34 of 594
Application +4 | -1
The eccentricity of the ellipse (x−1)29+(y+1)225=1\frac{(x-1)^2}{9}+\frac{(y+1)^2}{25}=1 is
View answer and explanation
Correct option: A
4/54 / 5
Option A (4 / 5) is correct because the eccentricity is found using the formula e=1−a2b2e = \sqrt{{1 - \frac{{a^2}}{{b^2}}}} where a2 = 9 and b2 = 25.
35 Easy
35 of 594
Application +4 | -1
The distance between the directrices of the ellipse x236+y220=1\frac{x^2}{36}+\frac{y^2}{20}=1 is
View answer and explanation
Correct option: C
18
Option C (1/31 / \sqrt{3}) is correct because the eccentricity is calculated using the given condition 2ae=6ae\frac{2a}{e} = 6ae.
36 Easy
36 of 594
Application +4 | -1
Angle between tangents drawn from the point (5,4)\left( {5,4} \right) to the ellipse x225+y216=1\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1, is
View answer and explanation
Correct option: B
900{90^0}
Option C ((3,4),(−13,4)(3,4),(-13,4)) is correct because these are the coordinates of the endpoints of the latus rectum of the given parabola after transforming it to standard form.
37 Easy
37 of 594
Application +4 | -1
The length of major and minor axis of an ellipse are 10 and 8 respectively and its major axis along y−axisy - axis the equation of the ellipse referred to its centre as origin is
View answer and explanation
Correct option: B
x216+y225=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{25}} = 1
Option B (x216+y225=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{25}} = 1) is correct because it represents the equation of the ellipse with the given major and minor axes along the y-axis.
38 Easy
38 of 594
Application +4 | -1
If a=[t2−3t+4]a=\left[t^2-3 t+4\right] and b=[3+5t]b=[3+5 t], where [.] denotes the greatest integer function, then the latusrectum of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 at t=32t=\frac{3}{2} is
View answer and explanation
Correct option: C
15\frac{1}{5}
Option D (110\frac{1}{10}) is correct because the latus rectum length is calculated using the greatest integer function and the given ellipse equation at t=3/2.
39 Easy
39 of 594
Application +4 | -1
The foci of the Basic Terms of Conics 25x2+16y2−150x=17525{x^2} + 16{y^2} - 150x = 175 are
View answer and explanation
Correct option: C
(3,±3)\left( {3, \pm 3} \right)
The correct answer is C ((3, ±3)) because completing the square and transforming the equation reveals a vertical ellipse centered at (3,0), and calculating the foci yields (3, ±3).
40 Medium
40 of 594
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.
41 Medium
41 of 594
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.
42 Easy
42 of 594
Understanding +4 | -1
The equation x=acos⁡θ,y=bsin⁡θ(a>b)x=a \cos \theta, y=b \sin \theta(a>b) represents a conic section whose eccentricity ee is given by
View answer and explanation
Correct option: C
e2=a2−b2a2e^2=\frac{a^2-b^2}{a^2}
Option C (e2=a2−b2a2e^2=\frac{a^2-b^2}{a^2}) is correct because the equation represents an ellipse, and the eccentricity of an ellipse is given by e=1−b2a2e = \sqrt{1 - \frac{b^2}{a^2}}, so e2=a2−b2a2e^2 = \frac{a^2 - b^2}{a^2}.
43 Medium
43 of 594
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.
44 Easy
44 of 594
Application +4 | -1
The locus of pole of tangents to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 with respect to the parabola y2=4axy^2=4 a x, is
View answer and explanation
Correct option: C
b2y2=4a2(x2−a2)b^2 y^2=4 a^2\left(x^2-a^2\right)
Option C (b2y2=4a2(x2−a2)b^2 y^2 = 4 a^2 (x^2 - a^2)) is correct because the locus of the pole is found by equating the polar equation to the tangent equation and using the condition for tangency.
45 Easy
45 of 594
Application +4 | -1
If the normal at any point PP on the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, meets the axes in GG and gg respectively, then PG:Pg=P G: P g=
View answer and explanation
Correct option: C
b2:a2b^2: a^2
Option C (b2:a2b^2: a^2) is correct because it represents the ratio of the distances PG and Pg after substituting the coordinates and simplifying.

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46 Easy
46 of 594
Application +4 | -1
If 2y=x2y = x and 3y+4x=03y + 4x = 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: C
13\sqrt {\frac{1}{3}}
Option C (13\sqrt {\frac{1}{3}}) is correct because it is derived from the relationship between the slopes of conjugate diameters and the eccentricity of the ellipse.
47 Easy
47 of 594
Understanding +4 | -1
The equation x210−a+y24−a=1\frac{x^2}{10-a}+\frac{y^2}{4-a}=1 represents an ellipse, if
View answer and explanation
Correct option: A
a<4a < 4
Option A (a<4a < 4) is correct because for the equation to represent an ellipse, both denominators must be positive, requiring aa to be less than 4.
48 Easy
48 of 594
Application +4 | -1
If ee is eccentricity of ellipse x2a2+y2b2=1(a>b)\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1(a > b) and e′e' is eccentricity of x2a2+y2b2=1(a<b),\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\left( {a < b} \right), then
View answer and explanation
Correct option: C
1e2+1(e′)2=1\frac{1}{{{e^2}}} + \frac{1}{{{{\left( {e'} \right)}^2}}} = 1
Option C (1e2+1(e′)2=1\frac{1}{{{e^2}}} + \frac{1}{{{{\left( {e'} \right)}^2}}} = 1) is correct because it represents the relationship between the eccentricities of an ellipse and its auxiliary ellipse.
49 Medium
49 of 594
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.
50 Easy
50 of 594
Application +4 | -1
The equation of the parabola whose vertex is at the centre of the ellipse x225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1 and the focus coincide with the focus of the ellipse on the positive side of the major axis of the ellipse is
View answer and explanation
Correct option: D
y2=12xy^2=12 x
Option D (y2=12xy^2=12 x) is correct because the focus of the ellipse is (3,0), which becomes the focus of the parabola, and the vertex is (0,0).
51 Easy
51 of 594
Application +4 | -1
The eccentricity of an ellipse whose centre is at the origin is 12\frac{1}{2}. If one directrix is x=−4x=-4, then the equation of the normal to it at (1,3/2)(1,3 / 2), is
View answer and explanation
Correct option: B
4x−2y=14 x-2 y=1
Option B (4x−2y=14 x-2 y=1) is correct because it is the equation of the normal to the ellipse at the given point.
52 Medium
52 of 594
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.
53 Easy
53 of 594
Application +4 | -1
In an ellipse, the distance between its foci is 6 and minor axis is 8 . The eccentricity, is
View answer and explanation
Correct option: D
3/53 / 5
Option D (3/53 / 5) is correct because using the given information about the foci and minor axis, we can calculate the eccentricity as e=3/5e = 3/5.
54 Easy
54 of 594
Understanding +4 | -1
If SS and S′S' are two foci of an ellipse x2a2+y2b2=1(a<b)\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1(a < b) and P(x1,y1)P\left( {{x_1},{y_1}} \right) a point on it, then SP+S′PSP + S'P is equal to
View answer and explanation
Correct option: B
2b2b
Option B (2b) is correct because the sum of the focal distances from any point on an ellipse is equal to the length of the major axis (2b) when a < b.
55 Easy
55 of 594
Application +4 | -1
If the eccentricity of an ellipse be 1/21 / \sqrt{2}, then its latus rectum is equal to its
View answer and explanation
Correct option: D
Semi-major axis
Option C (18) is correct because the distance between the directrices is calculated using the formula 2ae\frac{2a}{e}.
56 Easy
56 of 594
Analyzing +4 | -1
Let S={(x, y)∈R2:y21+r−x21−r=1}S = \left\{ {\left( {x,\,y} \right) \in {R^2}:\frac{{{y^2}}}{{1 + r}} - \frac{{{x^2}}}{{1 - r}} = 1} \right\}, where r≠±1r \ne \pm 1. Then, SS represents
View answer and explanation
Correct option: C
An ellipse whose eccentricity is 2r+1\sqrt {\frac{2}{{r + 1}}}, when r>1r > 1.
Option C (An ellipse whose eccentricity is 2r+1\sqrt {\frac{2}{{r + 1}}}, when r>1r > 1) is correct because when r > 1, the equation represents an ellipse, and its eccentricity is calculated as 2r+1\sqrt {\frac{2}{{r + 1}}}.
57 Easy
57 of 594
Application +4 | -1
The equation of tangent to the ellipse x2+4y2=5{x^2} + 4{y^2} = 5 at (−1,1)\left( { - 1,1} \right) is
View answer and explanation
Correct option: D
x−4y+5=0x - 4y + 5 = 0
Option D (x−4y+5=0x - 4y + 5 = 0) is correct because it represents the equation of the tangent to the ellipse at the specified point, obtained by direct substitution into the ellipse equation.
58 Easy
58 of 594
Application +4 | -1
The eccentricity of an ellipse is 13\frac{1}{3}, latus rectum is 4 and centre is (0,0)(0,0), then the equation of the ellipse is (given major axis is along xx-axis)
View answer and explanation
Correct option: B
16x281+2y29=1\frac{16 x^2}{81}+\frac{2 y^2}{9}=1
Option A (16x281+2y29=1\frac{16 x^2}{81}+\frac{2 y^2}{9}=1) is correct because it satisfies the given eccentricity and latus rectum conditions for the ellipse.
59 Easy
59 of 594
Application +4 | -1
If y=mx+cy=m x+c is the tangent on the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=1, then the value of cc is
View answer and explanation
Correct option: C
±9m2+4\pm \sqrt{9 m^2+4}
Option C (±9m2+4\pm \sqrt{{9 m^2+4}}) is correct because, for the ellipse x29+y24=1\frac{{x^2}}{{9}} + \frac{{y^2}}{{4}} = 1, the condition for tangency of the line y = mx + c is c=±a2m2+b2=±9m2+4c = \pm \sqrt{{a^2m^2 + b^2}} = \pm \sqrt{{9m^2 + 4}}.
60 Easy
60 of 594
Analyzing +4 | -1
The tangent at a point P(acos⁡θ,bsin⁡θ)P(a \cos \theta, b \sin \theta) on the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 meets the auxiliary circle in two points. The chord joining them subtends a right angle at the centre. Then, the eccentricity of the ellipse is given by
View answer and explanation
Correct option: A
(1+sin⁡2θ)−1/2\left(1+\sin ^2 \theta\right)^{-1 / 2}
Option A ((1+sin⁡2θ)−1/2\left(1+\sin ^2 \theta\right)^{-1 / 2}) is correct because it represents the eccentricity derived from the condition that the chord joining the intersection points subtends a right angle at the center.
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