Skip to main content
Back to ELLIPSE

Your Progress

0 / 594 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

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.

Take a chapter quiz →
Showing 61-90 of 594 questions

Preparing practice and saved progress…

61 Easy
61 of 594
Application +4 | -1
If for the ellipse x2α2+y2b2=1\frac{{{x^2}}}{{{\alpha ^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1, yy-axis is the minor axis and the length of the latusrectum is one half of the length of its minor axis, then its eccentricity is
View answer and explanation
Correct option: C
32\frac{{\sqrt 3 }}{2}
Option C (32\frac{{\sqrt 3 }}{2}) is correct because it is the eccentricity calculated from the given relationship between the latus rectum and minor axis lengths.
62 Easy
62 of 594
Application +4 | -1
For the ellipse 25x2+9y2−150x−90y+225=025{x^2} + 9{y^2} - 150x - 90y + 225 = 0, the eccentricity ee is equal to
View answer and explanation
Correct option: C
45\frac{4}{5}
Option C (45\frac{4}{5}) is correct because the eccentricity is calculated using the formula e=1−a2b2e = \sqrt{1 - \frac{a^2}{b^2}} after rewriting the ellipse equation.
63 Easy
63 of 594
Application +4 | -1
The locus of the mid-points of focal chords of the ellipse x2a2+y2b2=1,\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1, is
View answer and explanation
Correct option: B
x2a2+y2b2=exa2\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = \frac{{ex}}{{{a^2}}}
Option B (x2a2+y2b2=exa2\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = \frac{{ex}}{{{a^2}}}) is correct because it represents the locus of the midpoints of focal chords of the ellipse, derived by substituting the focus coordinates into the chord equation.
64 Medium
64 of 594
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.
65 Medium
65 of 594
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.
66 Medium
66 of 594
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.
67 Easy
67 of 594
Application +4 | -1
The number of values of c such that the line y=4x+cy = 4x + c touches the curve x24+y2=1\frac{{{x^2}}}{4} + {y^2} = 1 is
View answer and explanation
Correct option: B
2
Option B (2) is correct because there are two possible values of 'c' that satisfy the tangency condition for the ellipse.
68 Easy
68 of 594
Understanding +4 | -1
The number of maximum normals which can be drawn from a point to ellipse is
View answer and explanation
Correct option: A
4
The correct answer is A (4) because a maximum of four normals can be drawn from a point to an ellipse.
69 Medium
69 of 594
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.
70 Medium
70 of 594
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.
71 Medium
71 of 594
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.
72 Easy
72 of 594
Application +4 | -1
The equation of the ellipse with its centre at (1,2)(1,2), one focus at (6,2)(6,2) and passing through (4,6)(4,6), is
View answer and explanation
Correct option: A
(x−1)245+(y−2)220=1\frac{(x-1)^2}{45}+\frac{(y-2)^2}{20}=1
Option A ((x−1)245+(y−2)220=1\frac{(x-1)^2}{45}+\frac{(y-2)^2}{20}=1) is correct because it satisfies the given conditions: center at (1,2), focus at (6,2), and passing through (4,6).
73 Medium
73 of 594
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.
74 Medium
74 of 594
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.
75 Easy
75 of 594
Understanding +4 | -1
Which one of the following statements is not a false statement?
View answer and explanation
Correct option: B
qq : Circle is a particular case of an ellipse
The correct answer is 'Circle is a particular case of an ellipse' because a circle can be represented as an ellipse with equal major and minor axes.
76 Easy
76 of 594
Application +4 | -1
If PCP′P C P^{\prime} and DCD′D C D^{\prime} form a pair of conjugate diameters of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 and RR is any point on the circle x2+y2=c2x^2+y^2=c^2, then
View answer and explanation
Correct option: D
PR2+DR2+P′R2+D′R2=2(a2+b2+2c2)P R^2+D R^2+P^{\prime} R^2+D^{\prime} R^2=2\left(a^2+b^2+2 c^2\right)
Option D (PR2+DR2+P′R2+D′R2=2(a2+b2+2c2)P R^2 + D R^2 + P' R^2 + D' R^2 = 2(a^2 + b^2 + 2 c^2)) is correct because it represents the sum of squared distances from points on the ellipse and circle.
77 Medium
77 of 594
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.
78 Easy
78 of 594
Application +4 | -1
The eccentricity of the conic 36x2+144y2−36x−96y−119=036{x^2} + 144{y^2} - 36x - 96y - 119 = 0 is
View answer and explanation
Correct option: A
32\frac{{\sqrt 3 }}{2}
Option A (32\frac{{\sqrt 3 }}{2}) is correct because it's the eccentricity calculated after rewriting the conic equation in standard form and applying the eccentricity formula.
79 Medium
79 of 594
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.
80 Easy
80 of 594
Application +4 | -1
If the area of the auxiliary circle of the ellipse x2a2+y2b2=1(a>b)\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1(a > b) is twice the area of the ellipse, then the eccentricity of the ellipse is
View answer and explanation
Correct option: B
32\frac{{\sqrt 3 }}{2}
Option B (32\frac{{\sqrt 3 }}{2}) is correct because it is the calculated eccentricity of the ellipse given the area condition.
81 Easy
81 of 594
Application +4 | -1
Let e1e_1 and e2e_2 be the eccentricities of the ellipse x218+y24=1\frac{x^2}{18}+\frac{y^2}{4}=1 and hyperbola x29−y24=1\frac{x^2}{9}-\frac{y^2}{4}=1 respectively. If the point (e1,e2)\left(e_1, e_2\right) lies on the ellipse 15x2+3y2=k15 x^2+3 y^2=k, then the value of kk is
View answer and explanation
Correct option: C
16
Option C (16) is correct because it is the value obtained by substituting the eccentricities of the ellipse and hyperbola into the given equation.
82 Hard
82 of 594
Application +4 | -1
Tangent is drawn to the ellipse x227+y2=1\frac{{{x^2}}}{{27}} + {y^2} = 1 (33cos⁡θ,sin⁡θ)[where  θ∈(0,π2)]\left( {3\sqrt 3 \cos \mathrm{\theta },\sin \mathrm{\theta }} \right)\left[ {\mathrm{where}\,\,\mathrm{\theta } \in \left( {0,\frac{\pi }{2}} \right)} \right]. Then, the value of θ\theta such that sum of intercepts on axes made by this tangent is minimum, is
View answer and explanation
Correct option: B
π6\frac{\pi }{6}
Option B (π6\frac{\pi }{6}) is correct because the sum of the intercepts is minimized when tan⁡θ=13\tan \theta = \frac{1}{\sqrt 3 }, which occurs at θ=π6\theta = \frac{\pi }{6}.
83 Medium
83 of 594
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.
84 Easy
84 of 594
Application +4 | -1
If the line xcos⁡α+ysin⁡a=px\cos \alpha + y\sin \mathrm{a} = 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}({a^2}{\sec ^2}\alpha + {b^2}cose{c^2}\alpha ) = {\left( {{a^2} - {b^2}} \right)^2}
Option D (p2(a2sec⁡2α+b2cosec2α)=(a2−b2)2{p^2}({a^2}{\sec ^2}\alpha + {b^2}cose{c^2}\alpha ) = {\left( {{a^2} - {b^2}} \right)^2}) is correct because it represents the condition for the given line to be normal to the ellipse.
85 Easy
85 of 594
Application +4 | -1
The equation of the ellipse whose axes are along the coordinate axes, vertices are (±5,0)(\pm 5,0) are foci at (±4,0)(\pm 4,0), is
View answer and explanation
Correct option: B
9x2+25y2=2259 x^2+25 y^2=225
Option B (9x2+25y2=2259 x^2+25 y^2=225) is correct because it represents the equation of the ellipse with the given vertices and foci, derived by finding the values of a and b.
86 Easy
86 of 594
Application +4 | -1
If tangents drawn to the ellipse through point (1,23)(1,2 \sqrt{3}) to the ellipse x2/9+y2/b2=1x^2 / 9+y^2 / b^2=1 are at right angle, then value of bb is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the point must lie on the director circle of the ellipse, and substituting the point into the director circle equation allows us to solve for b.
87 Medium
87 of 594
Application +4 | -1
Length of major axis of 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 after converting to standard form, the length of the major axis is twice the value of b, where b is the semi-major axis.
88 Medium
88 of 594
Application +4 | -1
If the chords of constant 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 C (−a4b4- \frac{{{a^4}}}{{{b^4}}}) is correct because the product of the slopes of the chords of contact being -1 leads to the given relationship between the coordinates of the points.
89 Medium
89 of 594
Application +4 | -1
Equation of a tangent to the ellipse x236+y225=1\frac{x^2}{36}+\frac{y^2}{25}=1, passing through the point where a directrix of the ellipse meets the + ve xx-axis is
View answer and explanation
Correct option: C
6y+11x=366 y+\sqrt{11} x=36
Option C (6y+11x=366 y+\sqrt{11} x=36) is correct because it represents the tangent to the ellipse passing through the point where the directrix meets the positive x-axis.
90 Easy
90 of 594
Application +4 | -1
The eccentric angle of a point on the ellipse 9x2+4y2=369 x^2+4 y^2=36 at a distance 312\frac{\sqrt{31}}{2} units from the centre of the ellipse is &#39;θ\theta&#39;, then the value of ∣2cos⁡θ∣|2 \cos \theta| is-
View answer and explanation
Correct option: A
1
Option A (1) is correct. By substituting the parametric coordinates of a point on the ellipse and using the given distance from the center, the value of |2cosθ| can be determined.
Revision notes and key takeaways

Review ELLIPSE in your textbook or class notes, then use this set to check your understanding.

Showing 61-90 of 594 questions

Your practice

0 / 594 answered · 594 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.