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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Easy

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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.

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Showing 211-240 of 392 questions

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211 Easy
211 of 392
Application +4 | -1
For the ellipse 25x2+9y2−150x−90y+225=025 x^2+9 y^2-150 x-90 y+225=0, the eccentricity is
View answer and explanation
Correct option: C
4/54 / 5
Option C (4 / 5) is correct because, after rewriting the equation in standard form, the eccentricity is calculated using the formula e=1−a2b2e = \sqrt{{1 - \frac{{a^2}}{{b^2}}}} since b > a.
212 Easy
212 of 392
Application +4 | -1
Locus of the point of intersection of tangents at the end points of focal chord is
View answer and explanation
Correct option: A
x=ae′x=\frac{a}{e^\prime} if a>ba > b
Option A (x=ae′x=\frac{a}{e^{\prime}}, if a>ba > b) is correct because the locus of the intersection of tangents at the endpoints of a focal chord is the directrix of the ellipse.
213 Easy
213 of 392
Application +4 | -1
If x=6y+8x=6 y+8 be a tangent to the ellipse in standard form whose minor axis is 43\frac{4}{\sqrt{3}}, then its eccentricity is
View answer and explanation
Correct option: B
1112\sqrt{\frac{11}{12}}
Option B (1112\sqrt{\frac{11}{12}}) is correct because it is the eccentricity calculated using the tangent condition and the given minor axis length.
214 Easy
214 of 392
Application +4 | -1
The eccentricity of the ellipse represented by the equation 25  x2+16  y2−150  x−175=025\;{x^2} + 16\;{y^2} - 150\;x - 175 = 0 is
View answer and explanation
Correct option: B
3/53/5
Option B is correct because the eccentricity of the ellipse is calculated after rewriting the equation in standard form and using the relationship between a, b, and e.
215 Easy
215 of 392
Application +4 | -1
The eccentricity of the ellipse 4x2+y2−8x+2y+1=04 x^2+y^2-8 x+2 y+1=0 is
View answer and explanation
Correct option: B
3/2\sqrt{3} / 2
Option B (3/2\sqrt{3} / 2) is correct because the eccentricity is calculated using the formula e=1−a2b2e = \sqrt{{1 - \frac{{a^2}}{{b^2}}}} after rewriting the equation in standard ellipse form.
216 Easy
216 of 392
Application +4 | -1
The equation of the ellipse, whose one of the vertices is (0,7)(0,7) and the corresponding directrix is y=12y=12, is
View answer and explanation
Correct option: B
144x2+95y2=4655144 x^2+95 y^2=4655
Option B (144x2+95y2=4655144 x^2+95 y^2=4655) is the correct equation of the ellipse because it satisfies the given vertex and directrix conditions.
217 Easy
217 of 392
Analyzing +4 | -1
Let a line LL pass through the point intersection of the lines bx+10y−8=0b x+10 y-8=0 and 2x−3y=02 x-3 y=0, b∈R−{43}b \in R-\left\{\frac{4}{3}\right\}. If the line LL also passes through the point (1,1)(1,1) and touches the circle 17(x2+y2)=1617\left(x^2+y^2\right)=16, then the eccentricity of the ellipse x25+y25=1\frac{x^2}{5}+\frac{y^2}{5}=1 is
View answer and explanation
Correct option: B
35\sqrt{\frac{3}{5}}
Option D (35\sqrt{\frac{3}{5}}) is correct because after finding the equation of the line L and using the condition that it touches the circle, we can find the value of b and then calculate the eccentricity.
218 Easy
218 of 392
Remembering +4 | -1
If PP is any point on the ellipse x236+y216=1\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{16}} = 1 and SS and S′S' are the foci, then PS+PS′PS + PS' is equal to
View answer and explanation
Correct option: D
12
Option D (12) is correct because the sum of the focal distances of any point on the ellipse is equal to 2a, where a = 6 in this case.
219 Easy
219 of 392
Application +4 | -1
If polar of x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is always touching the circle x2+y2=c2x^2+y^2=c^2, then locus of pole is
View answer and explanation
Correct option: A
c2(a4y2+b4x2)=a4b4c^2\left(a^4 y^2+b^4 x^2\right)=a^4 b^4
Option B (c2(a4x2+b4y2)=a4b4c^2(a^4 x^2+b^4 y^2)=a^4 b^4) is correct because the polar of the ellipse touches the circle, so the condition for tangency can be used to find the locus of the pole.
220 Easy
220 of 392
Application +4 | -1
Coordinates of the foci of the ellipse 5x2+9y2+10x−36y−4=05{x^2} + 9{y^2} + 10x - 36y - 4 = 0 are
View answer and explanation
Correct option: A
(1,2)\left( {1,2} \right) and (−3,2)\left( { - 3,2} \right)
Option A ((1,21,2) and (−3,2- 3,2)) is correct because these are the coordinates of the foci after completing the square and finding the center and eccentricity.
221 Easy
221 of 392
Analyzing +4 | -1
A tangent to the ellipse 4x2+9y2=364 x^2+9 y^2=36 is cut by the tangent at the extremities of the major axis at TT and ' TT '. The circle on TT′T T^{\prime} as diameter passes through the point
View answer and explanation
Correct option: C
(0,0)(0,0)
Option C ((0,0)) is correct because the circle on TT′TT' as diameter passes through the origin.
222 Easy
222 of 392
Application +4 | -1
If the distance between the foci of an ellipse is half the length of its latus rectum, then the eccentricity of the ellipse is:
View answer and explanation
Correct option: B
2−1\sqrt{2}-1
Option B (2−1\sqrt{2}-1) is correct because the eccentricity is derived from the relationship between the distance between foci and the latus rectum length.
223 Easy
223 of 392
Application +4 | -1
The line lx+my−n=0l x+m y-n=0 is tangent to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 if
View answer and explanation
Correct option: A
a2l2+b2m2=n2a^2 l^2+b^2 m^2=n^2
Option A (a2l2+b2m2=n2a^2 l^2+b^2 m^2=n^2) is correct because this is the condition for the line lx+my−n=0lx + my - n = 0 to be tangent to the ellipse x2a2+y2b2=1\frac{{x^2}}{{a^2}} + \frac{{y^2}}{{b^2}} = 1.
224 Easy
224 of 392
Application +4 | -1
The locus of the point which moves such that the ratio of its distance from two fixed point in the plane is always a constant k(<1)k( < 1) is
View answer and explanation
Correct option: B
ellipse
Option B (ellipse) is correct because the locus of a point with a constant ratio of distances from two fixed points (less than 1) defines an ellipse.
225 Easy
225 of 392
Application +4 | -1
If P=(x,y),F1=(3,0),F2=(−3,0)P=(x, y), F_1=(3,0), F_2=(-3,0) and 16x2+25y2=40016 x^2+25 y^2=400, then PF1+PF1P F_1+P F_1 equals:
View answer and explanation
Correct option: C
10
Option C (10) is correct because PF1 + PF2 equals the length of the major axis, which is 2a, and a is 5 in this ellipse equation.
226 Easy
226 of 392
Application +4 | -1
The equation of the ellipse (referred to its axes as the axes of xx and yy respectively) which passes through the point (−3,1)\left( { - 3,1} \right) and has eccentricity 25,\sqrt {\frac{2}{5}} , is
View answer and explanation
Correct option: C
3x2+5y2=323{x^2} + 5{y^2} = 32
Option C (3x2+5y2=323{x^2} + 5{y^2} = 32) is correct because it represents the equation of an ellipse that passes through the point (-3, 1) and has the given eccentricity.
227 Easy
227 of 392
Application +4 | -1
The line lx+my+n=0l x+m y+n=0 is a normal to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, if
View answer and explanation
Correct option: B
a2l2+b2m2=(a2−b2)2n2\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{\left(a^2-b^2\right)^2}{n^2}
Option B (a2l2+b2m2=(a2−b2)2n2\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{\left(a^2-b^2\right)^2}{n^2}) is correct because it represents the condition for the line to be normal to the ellipse.
228 Easy
228 of 392
Application +4 | -1
The length of latusrectum of the ellipse 9x2+16y2=1449{x^2} + 16{y^2} = 144 is
View answer and explanation
Correct option: D
92\frac{9}{2}
Option D (9/2) is correct because the length of the latus rectum of the ellipse is calculated using the formula 2b²/a, where a and b are the semi-major and semi-minor axes, respectively.
229 Easy
229 of 392
Application +4 | -1
For the ellipse, 4(x−2y+1)2+9(2x+y+2)2=180,4{(x - 2y + 1)^2} + 9{(2x + y + 2)^2} = 180, lengths of major and minor axes are respectively.
View answer and explanation
Correct option: C
6 and 4
Option A (656 \sqrt{5} and 454 \sqrt{5}) is incorrect. The correct lengths of the major and minor axes are 6 and 4, respectively, obtained by simplifying the ellipse equation.
230 Easy
230 of 392
Application +4 | -1
If the eccentricities of the ellipse x24+y23=1\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1 and the hyperbola x264−y2b2=1\frac{{{x^2}}}{{64}} - \frac{{{y^2}}}{{{b^2}}} = 1 are reciprocals of each other, then b2{b^2} is equal to
View answer and explanation
Correct option: A
192
Option A (192) is correct because the eccentricities are reciprocals, leading to the equation 1 + b2/64 = 4, which solves to b2 = 192.
231 Easy
231 of 392
Application +4 | -1
The eccentricity of the ellipse 25x2+16y2−150x−175=025 x^2+16 y^2-150 x-175=0 is
View answer and explanation
Correct option: D
3/53 / 5
Option D (3/53 / 5) is correct because it is the calculated eccentricity of the given ellipse after converting it to standard form.
232 Easy
232 of 392
Application +4 | -1
The dimensions of the rectangle of maximum area that can be inscribed in the ellipse (x/4)2+(y/3)2(x / 4)^2+(y / 3)^2 =1=1 are
View answer and explanation
Correct option: C
28,322 \sqrt{8}, 3 \sqrt{2}
Option C (28,322 \sqrt{8}, 3 \sqrt{2}) is correct because these are the dimensions of the rectangle of maximum area inscribed in the given ellipse.
233 Easy
233 of 392
Application +4 | -1
The equation of the ellipse whose distance between foci is equal to 8 and distance between the directrix is 18, is
View answer and explanation
Correct option: D
5x2+9y2=1805{x^2} + 9{y^2} = 180
Option D (5x2+9y2=1805{x^2} + 9{y^2} = 180) is correct because it represents the ellipse with the given distance between foci and directrices.
234 Easy
234 of 392
Application +4 | -1
Equation of a line joining a foci of the ellipse x225+y29=1\frac{x^2}{25}+\frac{y^2}{9}=1 to a foci of the ellipse x29+y225=1\frac{x^2}{9}+\frac{y^2}{25}=1 is,
View answer and explanation
Correct option: A
x+y=4x+y=4
Option A (x+y=4x+y=4) is correct because it represents the line joining the foci of the two given ellipses.
235 Easy
235 of 392
Application +4 | -1
If y=x+1y=x+1 is a polar to x2+2y2=1x^2+2 y^2=1 then pole is
View answer and explanation
Correct option: A
(−1,12)\left(-1, \frac{1}{2}\right)
Option A ((−1,12)\left(-1, \frac{1}{2}\right)) is correct because it is found by comparing the given polar equation with the general polar equation T=0.
236 Easy
236 of 392
Application +4 | -1
The locus of the point of intersection of perpendicular tangents to x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 and x2a2+λ+y2b2+λ=1\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1, is
View answer and explanation
Correct option: C
x2+y2=a2+b2+λx^2+y^2=a^2+b^2+\lambda
Option C (x2+y2=a2+b2+λx^2+y^2=a^2+b^2+\lambda) is correct because it represents the locus of the intersection point of perpendicular tangents to the two given ellipses.
237 Easy
237 of 392
Application +4 | -1
An ellipse is drawn by taking a diameter of the circle (x−1)2+y2=1(x-1)^2+y^2=1 as its semi-minor axis and diameter of the circle x2+(y−2)2=4x^2+(y-2)^2=4 is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is
View answer and explanation
Correct option: D
x2+4y2=16x^2+4 y^2=16
Option D (x2+4y2=16x^2+4 y^2=16) is correct because it correctly uses the diameters of the given circles as the semi-major and semi-minor axes of the ellipse.
238 Easy
238 of 392
Application +4 | -1
If BB and B′B' are the ends of minor axis and SS and S′S' are the foci of the ellipse x225+y29=1,\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1, then area of the rhombus SBS′B′SBS'B' will be
View answer and explanation
Correct option: C
24 sq. units
Option C (24 sq. units) is correct because the area of the rhombus is half the product of its diagonals, which are the minor axis and the distance between the foci.
239 Easy
239 of 392
Application +4 | -1
If the length of the major axis of an ellipse is three times the length of its minor axis, its eccentricity, is
View answer and explanation
Correct option: D
223\frac{{2\sqrt 2 }}{3}
Option D (223\frac{{2\sqrt 2 }}{3}) is correct because it's the eccentricity calculated from the relationship between the major and minor axes of the ellipse.
240 Easy
240 of 392
Application +4 | -1
The eccentricity of the ellipse 9x2+5y2−30y=09 x^2+5 y^2-30 y=0 is
View answer and explanation
Correct option: B
2/32 / 3
Option B (2 / 3) is correct because the eccentricity is calculated using the formula e=1−a2b2e = \sqrt{{1 - \frac{{a^2}}{{b^2}}}} after rewriting the equation in standard ellipse form.
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