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 241-270 of 594 questions

Preparing practice and saved progress…

241 Easy
241 of 594
Application +4 | -1
Let LL′LL' be the latusrectum and S be a focus of the ellipse x2a2+y2b2=1.\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1. If ΔSLL′\Delta SLL' is equilaterial, 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 when the triangle SLL' is equilateral, the eccentricity can be derived using the properties of the ellipse and the latus rectum.
242 Easy
242 of 594
Application +4 | -1
Let the equations of two ellipses be
E1:x23+y22=1 and E2:x216+y2b2=1,E_{1}: \frac{x^{2}}{3}+\frac{y^{2}}{2}=1 \text { and } E_{2}: \frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1,

If the product of their eccentricities is 12\frac{1}{2}, then the length of theminor axis of ellipse E2E_{2} is :
View answer and explanation
Correct option: C
4
Option C (4) is correct because the length of the minor axis of ellipse E2E_2 is calculated using the product of eccentricities and the given equation.
243 Medium
243 of 594
Application +4 | -1
The locus of points whose polars with respect to the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 are at a distance dd from the centre of the ellipse, is
View answer and explanation
Correct option: B
x2a4+y2b4=1d2\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} = \frac{1}{{{d^2}}}
Option B (x2a4+y2b4=1d2\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} = \frac{1}{{{d^2}}}) is correct because it represents the locus of points whose polars are at a distance d from the center of the ellipse.
244 Easy
244 of 594
Application +4 | -1
The foci of an ellipse are (0,±4)\left( {0, \pm 4} \right) and the equations for the directrices are y=±9y = \pm 9. The equation for the ellipse is
View answer and explanation
Correct option: D
9x2+5y2=1809{x^2} + 5{y^2} = 180
Option D (9x2+5y2=1809{x^2} + 5{y^2} = 180) is correct because it represents the equation of the ellipse with the given foci and directrices.
245 Medium
245 of 594
Application +4 | -1
If roots of quadratic equation ax2+2bx+c=0a{x^2} + 2bx + c = 0 are not real, then ax2+2bxy+cy2+dx+ey+f=0a{x^2} + 2bxy + c{y^2} + dx + ey + f = 0 represents a/an
View answer and explanation
Correct option: A
Ellipse
Option A (Ellipse) is correct because if the roots of the quadratic equation are not real, then b2<acb^2 < ac, which implies the given equation represents an ellipse.
246 Easy
246 of 594
Application +4 | -1
The locus of midpoints of a focal chord 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
x2a2+y2b2=±exa\frac{x^2}{a^2}+\frac{y^2}{b^2}=\pm \frac{e x}{a}
Option A (x2a2+y2b2=±exa\frac{{x^2}}{{a^2}}+\frac{{y^2}}{{b^2}}=\pm \frac{{e x}}{{a}}) is correct because the locus of the midpoints of a focal chord is obtained by using the equation of the chord with a given midpoint and the fact that the chord passes through the focus.
247 Easy
247 of 594
Application +4 | -1
The acute angle between the pair of tangents drawn to the ellipse 2x2+3y2=52 x^2+3 y^2=5 from the point (1,3)(1,3) is
View answer and explanation
Correct option: B
tan⁡−1(2475)\tan ^{-1}\left(\frac{24}{7 \sqrt{5}}\right)
Option B is correct because the acute angle between the tangents is found using the formula involving the slopes of the tangents and the given ellipse equation.
248 Easy
248 of 594
Application +4 | -1
The equation of the ellipse passing through (2,1) having e=1/2e = 1/2 is
View answer and explanation
Correct option: A
3x2+4y2=163{x^2} + 4{y^2} = 16
Option A (3x2+4y2=163{x^2} + 4{y^2} = 16) is correct because it satisfies the ellipse equation given the eccentricity and the point (2,1).
249 Easy
249 of 594
Application +4 | -1
If pair of tangents drawn to the ellipse x216+y29=1\frac{x^2}{16}+\frac{y^2}{9}=1 from a point PP, so that angle between the tangents are at right angle, then possible coordinates of the point PP is/are
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all the points satisfy the equation of the director circle, which is the locus of points from which perpendicular tangents can be drawn.
250 Easy
250 of 594
Application +4 | -1
The equation of the line passing through the centre and bisecting the chord 7x+y−1=07 x+y-1=0 of the ellipse x21+y27=1\frac{x^2}{1}+\frac{y^2}{7}=1 is
View answer and explanation
Correct option: A
x=yx=y
Option A (x=yx=y) is correct because the line passing through the center and bisecting the chord of the ellipse has the same slope as the line connecting the origin to the midpoint of the chord.
251 Medium
251 of 594
Application +4 | -1
The eccentricity of the ellipse 25x2+16y2=40025 x^2+16 y^2=400 is
View answer and explanation
Correct option: A
3/53 / 5
Option A (3/53 / 5) is correct because the eccentricity is calculated using the formula e=1−a2b2e = \sqrt{{1 - \frac{{a^2}}{{b^2}}}} since b>ab > a.
252 Easy
252 of 594
Application +4 | -1
The eccentricity of the ellipse 4x2+9y2+8x+36y+4=04 x^2+9 y^2+8 x+36 y+4=0 is
View answer and explanation
Correct option: D
5/3\sqrt{5} / 3
Option D (5/3\sqrt{5} / 3) is correct because the eccentricity is calculated using the formula e=1−b2a2e = \sqrt{{1 - \frac{{b^2}}{{a^2}}}} after rewriting the equation in standard ellipse form.
253 Easy
253 of 594
Application +4 | -1
If chords of contact of the tangent from two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 are at right angles, then x1x2a2×y1y2b2\frac{x_1x_2}{a^2} \times \frac{y_1y_2}{b^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 must equal -1 due to their perpendicularity.
254 Medium
254 of 594
Application +4 | -1
Let PiP_i and Pi′P_i^{\prime} be the feet of the perpendiculars drawn from the foci SS and S′S^{\prime} on a tangent TiT_i to an ellipse whose length of semi-major axis is 20 . If ∑i=110(SPi)(S′Pi′)=2560\sum\limits_{{i=1}}^{10}\left(S P_i\right)\left(S^{\prime} P_i^{\prime}\right)=2560, then the value of eccentricity is
View answer and explanation
Correct option: C
35\frac{3}{5}
Option C (35\frac{3}{5}) is correct because the eccentricity is calculated using the relationship between the semi-major axis, semi-minor axis, and eccentricity of the ellipse.
255 Easy
255 of 594
Remembering +4 | -1
The sum of focal distance of any point on the ellipse with major and minor axes as 2a2a and 2b2b respectively, is equal to
View answer and explanation
Correct option: A
2a2a
Option A (2a2a) is correct because the sum of the focal distances of any point on the ellipse is equal to the length of the major axis, which is 2a.

Explore Other Chapters in MATHEMATICS

View All →
SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
256 Medium
256 of 594
Application +4 | -1
Chords of an ellipse are drawn through the positive end of the minor axis. Then, their mid-point lies on
View answer and explanation
Correct option: C
an ellipse
Option C (an ellipse) is correct because the locus of the mid-point of the chords is found to be an ellipse by using the equation of the chord and the condition that it passes through the end of the minor axis.
257 Easy
257 of 594
Application +4 | -1
The eccentricity of conic 4(2y−x−3)2+9(2x+y−1)2=4(2 y-x-3)^2+9(2 x+y-1)^2= 80 is
View answer and explanation
Correct option: B
53\frac{\sqrt{5}}{3}
Option B (53\frac{{\sqrt{{5}}}}{{3}}) is correct because it represents the eccentricity of the ellipse after transforming the given equation into standard form.
258 Easy
258 of 594
Application +4 | -1
The parametric representation of a point of the ellipse whose foci are (3, 0) and (−1,0)\left( { - 1,0} \right) and eccentricity 23\frac{2}{3} is
View answer and explanation
Correct option: D
(1+3cosθ,5sin⁡θ)(1 + 3\mathrm{cos \theta },\sqrt 5 \sin \mathrm{\theta })
The correct answer is D ((1+3cosθ,5sin⁡θ)(1 + 3\mathrm{cos \theta },\sqrt 5 \sin \mathrm{\theta })) because it represents the parametric coordinates of a point on the ellipse with the given foci and eccentricity.
259 Easy
259 of 594
Analyzing +4 | -1
Let E1:x2a2+y2b2=1,  a>b{E_1}:\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\,\,a > b. Let E2{E_2} be another ellipse such that it touches the end points of major axis of E1{E_1} and the foci of E2{E_2} are the end points of minor axis of E1{E_1}. If E1{E_1} and E2{E_2} have same eccentricities, then its value is
View answer and explanation
Correct option: A
−1+52\frac{{ - 1 + \sqrt 5 }}{2}
Option A (−1+52\frac{{ - 1 + \sqrt 5 }}{2}) is correct because the eccentricity is derived from the geometric relationships between the two ellipses and solving the resulting quadratic equation.
260 Medium
260 of 594
Application +4 | -1
If the tangents on the ellipse 4x2+y2=84{x^2} + {y^2} = 8 at the points (1, 2) and (a, b)(a,\,b) are perpendicular to each other, then a2{a^2} is equal to
View answer and explanation
Correct option: D
217\frac{2}{{17}}
Option D (217\frac{{2}}{{17}}) is correct because it is the value of a2{a^2} obtained by using the perpendicularity condition of tangents and the equation of the ellipse.
261 Easy
261 of 594
Understanding +4 | -1
If PP is a point on the ellipse x216+y220=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{20}} = 1 whose foci are SS and S′S'. Then PS+PS′PS + PS' is
View answer and explanation
Correct option: B
454\sqrt 5
Option B (4√5) is correct because the sum of the distances from any point on the ellipse to its foci is equal to the length of the major axis, which is 2a = 2√(20) = 4√5.
262 Easy
262 of 594
Analyzing +4 | -1
Locus of the middle points of all chords of x24+y29=1\frac{x^2}{4}+\frac{y^2}{9}=1, which are at a distance of 2 units from the vertex of parabola y2=−8axy^2=-8ax, is
View answer and explanation
Correct option: B
(x24+y29)2=4(x216+y281)\left(\frac{x^2}{4}+\frac{y^2}{9}\right)^2=4\left(\frac{x^2}{16}+\frac{y^2}{81}\right)
Option B ((x24+y29)2=4(x216+y281)\left(\frac{x^2}{4}+\frac{y^2}{9}\right)^2=4\left(\frac{x^2}{16}+\frac{y^2}{81}\right)) is correct because it represents the locus of the midpoints of the chords at a given distance from the vertex of the parabola.
263 Easy
263 of 594
Analyzing +4 | -1
If α−β=\alpha-\beta= constant, then the locus of the point of intersection of tangents at P(acos⁡α,bsin⁡α)P(a \cos \alpha, b \sin \alpha) and Q(acos⁡β,bsin⁡β)Q(a \cos \beta, b \sin \beta) to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, is
View answer and explanation
Correct option: C
An ellipse
Option C (An ellipse) is correct because the locus of the point of intersection of tangents at P and Q to the ellipse is another ellipse.
264 Easy
264 of 594
Application +4 | -1
If f(x)f(x) is a decreasing function, then the set of values of 'kk', for which the major axis of the ellipse x2f(k2+2k+5)+y2f(k+11)=1\frac{x^2}{f\left(k^2+2 k+5\right)}+\frac{y^2}{f(k+11)}=1 is the XX-axis, is
View answer and explanation
Correct option: B
k∈(−3,2)k \in(-3,2)
Option B (k∈(−3,2)k \in(-3,2)) is correct because it represents the interval where the inequality k2+k−6<0k^2 + k - 6 < 0 holds true, ensuring that f(k2+2k+5)>f(k+11)f(k^2+2k+5) > f(k+11) for a decreasing function f(x).
265 Easy
265 of 594
Application +4 | -1
Let PP be the variable point on the ellipse x225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1 with focii at S&S′S \& S^{\prime}. Then maximum area of △PSS′\triangle \mathrm{PSS}^{\prime}
View answer and explanation
Correct option: B
12
Option B (12) is correct because the maximum area of the triangle is given by abe, where e is the eccentricity.
266 Easy
266 of 594
Application +4 | -1
If a bar of given length moves with its extremities on two fixed straight lines at right angles, then the locus of any point on bar marked on the bar describes a/an
View answer and explanation
Correct option: C
Ellipse
Option C (Ellipse) is correct because the locus of a point on a bar moving with its extremities on two fixed perpendicular lines is an ellipse.
267 Medium
267 of 594
Analyzing +4 | -1
The length of the major axis of the ellipse (5x−10)2+(5y+15)2=(3x−4y+7)24(5 x-10)^2+(5 y+15)^2=\frac{(3 x-4 y+7)^2}{4} is
View answer and explanation
Correct option: B
203\frac{20}{3}
Option B (203\frac{20}{3}) is correct because it is calculated by finding the distance from the focus to the directrix and using the eccentricity to find the length of the major axis.
268 Easy
268 of 594
Application +4 | -1
The locus of the middle point of the portion of a tangent to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 included between the axes is the curve
View answer and explanation
Correct option: B
a2x2+b2y2=4\frac{a^2}{x^2}+\frac{b^2}{y^2}=4
Option B (a2x2+b2y2=4\frac{{a^2}}{{x^2}}+\frac{{b^2}}{{y^2}}=4) is correct because it represents the locus of the midpoint of the tangent intercepted between the axes.
269 Easy
269 of 594
Application +4 | -1
If (−4,3)\left( { - 4,3} \right) and (8,3)\left( {8,3} \right) are the vertices of an ellipse whose eccentricity is 5/65/6 then the equation of the ellipse is
View answer and explanation
Correct option: B
(x−2)236+(y−3)211=1\frac{{{{\left( {x - 2} \right)}^2}}}{{36}} + \frac{{{{\left( {y - 3} \right)}^2}}}{{11}} = 1
Option B ((x−2)236+(y−3)211=1\frac{{{{\left( {x - 2} \right)}^2}}}{{36}} + \frac{{{{\left( {y - 3} \right)}^2}}}{{11}} = 1) is correct because it is the equation of the ellipse with the given vertices and eccentricity.
270 Easy
270 of 594
Application +4 | -1
Length of a latera recta of the ellipse x281+y263=1\frac{x^2}{81}+\frac{y^2}{63}=1 is (in units)
View answer and explanation
Correct option: C
14
Option C (14) is correct because the length of the latus rectum of an ellipse is given by 2b2a\frac{{2b^2}}{{a}}, and substituting the values gives 14.
Revision notes and key takeaways

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

Showing 241-270 of 594 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - ELLIPSE MCQs (Easy) 392 MCQs → ENGINEERING MATHEMATICS - ELLIPSE MCQs (Hard) 15 MCQs → ENGINEERING MATHEMATICS - ELLIPSE MCQs (Medium) 187 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ SEQUENCE OF SERIES 1764 MCQs → DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → INDEFINITE INTEGRATION 1179 MCQs

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

Practice settings

Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

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.

Continue with Free Account →