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

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

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 1-30 of 594 questions
1 Medium
1 of 594
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 Easy
2 of 594
Application +4 | -1
Let PP be a variable point on the ellipse x225+y216=1\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1 with foci at SS and S′.S'. If AA be the area of triangle PSS′,PSS', then the maximum value of AA is
View answer and explanation
Correct option: B
12 sq. units
Option B (12 sq. units) is correct because the maximum area of the triangle PSS' is half the product of the distance between the foci and the semi-minor axis.
3 Medium
3 of 594
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.
4 Easy
4 of 594
Application +4 | -1
The slope of a common tangent to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 and a concentric circle of radius rr, is
View answer and explanation
Correct option: B
r2−b2a2−r2\sqrt{\frac{r^2-b^2}{a^2-r^2}}
Option B (r2−b2a2−r2\sqrt{\frac{r^2-b^2}{a^2-r^2}}) is correct because equating the condition for tangency of a line to both the ellipse and the circle yields this slope.
5 Easy
5 of 594
Analyzing +4 | -1
A tangent to the ellipse x225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1 at any point PP meet the line x=0x=0 at a point QQ. Let RR be the image of QQ in the line y=xy=x, then circle whose extremities of a diameter are QQ and RR passes through a fixed point. The fixed point is
View answer and explanation
Correct option: D
(0,0)(0,0)
Option D ((0,0)) is correct because the circle formed with diameter QR always passes through the origin, regardless of the point P on the ellipse.
6 Medium
6 of 594
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.
7 Medium
7 of 594
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.
8 Hard
8 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 ends BB and B′B' of minor axis at LL and L′L' respectively, then BL.B′L′=BL.B'L' =
View answer and explanation
Correct option: A
a2{a^2}
Option A (a2{a^2}) is correct because it is the product of the distances from the points where the tangent at any point P on the ellipse meets the tangents at the ends of the minor axis.
9 Medium
9 of 594
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.
10 Easy
10 of 594
Analyzing +4 | -1
The point at shortest distance from the line x+y=7x+y=7 and lying on an ellipse x2+2y2=6x^2+2 y^2=6, has coordinates
View answer and explanation
Correct option: C
(2,1)(2,1)
Option C ((2,1)) is correct because the tangent to the ellipse at this point is parallel to the given line, indicating the shortest distance.
11 Medium
11 of 594
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.
12 Medium
12 of 594
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.
13 Easy
13 of 594
Application +4 | -1
For the ellipse x264+y228=1\frac{x^2}{64}+\frac{y^2}{28}=1, the eccentricity is
View answer and explanation
Correct option: A
34\frac{3}{4}
Option A (34\frac{{3}}{{4}}) is correct because the eccentricity is calculated using the formula e=1−b2a2e = \sqrt{{1 - \frac{{b^2}}{{a^2}}}} with the given values of a and b.
14 Medium
14 of 594
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.
15 Medium
15 of 594
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.

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
16 Easy
16 of 594
Understanding +4 | -1
If equation of the ellipse is x2100+y2400=1\frac{x^2}{100}+\frac{y^2}{400}=1, then

I. Vertices of the ellipse are (0,±20)(0, \pm 20)

II. Foci of the ellipse are (0,±103)(0, \pm 10 \sqrt{3})

III. Length of major axis is 40

IV. Eccentricity of the ellipse is 32\frac{\sqrt{3}}{2}.
View answer and explanation
Correct option: D
All are true
Option D is correct because all the statements (I, II, III, and IV) are true regarding the given ellipse equation.
17 Hard
17 of 594
Application +4 | -1
The locus of the poles of normal 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: C
a6x2+b6y2=(a2−b2)2\frac{{{a^6}}}{{{x^2}}} + \frac{{{b^6}}}{{{y^2}}} = {\left( {{a^2} - {b^2}} \right)^2}
Option C (a6x2+b6y2=(a2−b2)2\frac{{{a^6}}}{{{x^2}}} + \frac{{{b^6}}}{{{y^2}}} = {\left( {{a^2} - {b^2}} \right)^2}) is correct because it represents the locus of the poles of normal chords of the ellipse, derived by finding the equation of the normal and the polar, then eliminating the parameter.
18 Medium
18 of 594
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.
19 Easy
19 of 594
Application +4 | -1
If α,β\alpha, \beta are the eccentric angles of the extremities of a focal chord of the ellipse x2/16+y2/9=1x^2 / 16+y^2 / 9=1, then tan⁡(α/2)tan⁡(β/2)=\tan (\alpha / 2) \tan (\beta / 2)=
View answer and explanation
Correct option: D
87−239\frac{8 \sqrt{7}-23}{9}
Option D (87−239\frac{8 \sqrt{7}-23}{9}) is correct because it represents the value of tan⁡(α/2)tan⁡(β/2)\tan(\alpha/2) \tan(\beta/2) for a focal chord of the ellipse.
20 Easy
20 of 594
Application +4 | -1
The total number of tangents through the points (3,5)\left( {3,5} \right) that can be drawn to the ellipses 3x2+5y2=323{x^2} + 5{y^2} = 32 and 25x2+9y2=45025{x^2} + 9{y^2} = 450 is
View answer and explanation
Correct option: C
3
Option A (3) is correct because the point (3,5) lies outside the first ellipse (2 tangents) and on the second ellipse (1 tangent).
21 Hard
21 of 594
Analyzing +4 | -1
The curve represented by the equation x2sin⁡2−cos⁡3+y2sin⁡3−cos⁡2=1\frac{x^2}{\sin \sqrt{2}-\cos \sqrt{3}}+\frac{y^2}{\sin \sqrt{3}-\cos \sqrt{2}}=1 is
View answer and explanation
Correct option: A
An ellipse with foci on XX-axis
Option A (An ellipse with foci on XX-axis) is correct because both denominators are positive and the denominator of x2x^2 is greater than the denominator of y2y^2.
22 Medium
22 of 594
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.
23 Medium
23 of 594
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.
24 Medium
24 of 594
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).
25 Easy
25 of 594
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 major axis is
View answer and explanation
Correct option: A
12\frac{1}{{\sqrt 2 }}
Option A (12\frac{1}{{\sqrt 2 }}) is correct because the eccentricity is derived from the relationship between the latus rectum and the major axis.
26 Easy
26 of 594
Application +4 | -1
The eccentricity of the ellipse ax2+by2+2fxa x^2+b y^2+2 f x +2gy+c=0+2 g y+c=0 if axis of ellipse parallel to XX-axis is
View answer and explanation
Correct option: A
(b−ab)\sqrt{\left(\frac{b-a}{b}\right)}
Option A ((b−ab)\sqrt{\left(\frac{{b-a}}{{b}}\right)}) is correct because it is the standard formula for the eccentricity of an ellipse with its major axis parallel to the x-axis.
27 Medium
27 of 594
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.
28 Easy
28 of 594
Application +4 | -1
Equation of a directrix of the ellipse x236+y24=1\frac{x^2}{36}+\frac{y^2}{4}=1 is
View answer and explanation
Correct option: C
2x+9=0\sqrt{2} x+9=0
Option C (2x+9=0\sqrt{2} x+9=0) is correct because it represents the equation of a directrix of the given ellipse.
29 Easy
29 of 594
Application +4 | -1
The centre of the ellipse 4x2+9y2−16x−54y+61=04 x^2+9 y^2-16 x-54 y+61=0 is
View answer and explanation
Correct option: C
(3,2)(3,2)
Option B ((2,3)) is correct because completing the square allows us to rewrite the equation in standard ellipse form, revealing the center.
30 Easy
30 of 594
Understanding +4 | -1
An ellipse is the set of all points in a plane, the sum of whose
distances from two fixed points in the plane is a constant.

I. The two fixed points are called the foci of the ellipse.

II. The mid point of the line segment joining the foci is
called the centre of the ellipse.

III. The end points of the major axis are called the vertices
of the ellipse.
View answer and explanation
Correct option: D
All are correct
Option D is correct because all the statements (I, II, and III) are correct definitions related to an ellipse.
Revision notes and key takeaways

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

Key Takeaways

  • ✓ This set contains 594 single-correct-answer MCQs on ELLIPSE.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
Showing 1-30 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
📚 Switch Set (ALL)
Chapter Sets
ALL 594 MCQs EASY 392 MCQs HARD 15 MCQs MEDIUM 187 MCQs
0 /594
0✓ 0✗
0% Accuracy
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 →