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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 181-210 of 392 questions

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181 Easy
181 of 392
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.
182 Easy
182 of 392
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.
183 Easy
183 of 392
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.
184 Easy
184 of 392
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.
185 Easy
185 of 392
Application +4 | -1
If the normal at the point P(θ)P(\theta) to the ellipse x214+y25=1\frac{x^2}{14}+\frac{y^2}{5}=1 intersects it again at the point Q(2θ)Q(2 \theta), 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 is the solution to the equation derived from the intersection of the normal and the ellipse.
186 Easy
186 of 392
Application +4 | -1
The eccentricity of the ellipse 9x2+5y2−18x−2y−16=09 x^2+5 y^2-18 x-2 y-16=0 is
View answer and explanation
Correct option: B
2/32 / 3
Option B (2 / 3) is correct because, after rewriting the equation in standard form, the eccentricity is calculated using the formula e=b2−a2b2e = \sqrt{{\frac{{b^2 - a^2}}{{b^2}}}} since b > a.
187 Easy
187 of 392
Application +4 | -1
The equation of the ellipse whose axes are of length 6 and 262 \sqrt{6} and their equations are x−3y+3=0x-3 y+3=0 and 3x+y−1=03 x+y-1=0 respectively, is
View answer and explanation
Correct option: B
3(x−3y+3)2+2(3x+y−1)2=1803(x-3 y+3)^2+2(3 x+y-1)^2=180
Option B (3(x−3y+3)2+2(3x+y−1)2=1803(x-3 y+3)^2+2(3 x+y-1)^2=180) is correct because it represents the equation of the ellipse with given axes lengths and equations.
188 Easy
188 of 392
Application +4 | -1
The eccentricity of the conic (x+2)27+(y−1)2=14\frac{{{{\left( {x + 2} \right)}^2}}}{7} + {\left( {y - 1} \right)^2} = 14 is
View answer and explanation
Correct option: D
67\sqrt {\frac{6}{7}}
The correct answer is 67\sqrt {\frac{6}{7}} because it is the eccentricity calculated from the given equation of the ellipse.
189 Easy
189 of 392
Application +4 | -1
ABAB is a diameter of x2+9y2=25.{x^2} + 9{y^2} = 25. The eccentric angle of AA is π/6\pi /6. Then, the eccentric angle of BB is
View answer and explanation
Correct option: B
−5 π/6- 5\,\pi /6
Option B ( −5,π/6\ - 5\\,\pi /6) is correct because it represents the eccentric angle of point B, which is diametrically opposite to point A.
190 Easy
190 of 392
Application +4 | -1
The equation of a directrix of the ellipse x216+y225=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{25}} = 1 is
View answer and explanation
Correct option: C
3y=±253y = \pm 25
The correct answer is C (3y=±253y = \pm 25) because it represents the equation of the directrix of the given ellipse.
191 Easy
191 of 392
Application +4 | -1
Consider an ellipse with major and minor axes of length 10 and 8 units respectively. The radius of largest circle that can be inscribed in this ellipse, it is given that centre of this circle is one focus of the ellipse is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the radius of the largest circle that can be inscribed in the ellipse with its center at one focus is equal to a - ae, where a is the semi-major axis and e is the eccentricity.
192 Easy
192 of 392
Application +4 | -1
The equation of chord of ellipse 2x2+5y2=202x^2+5y^2=20 which is bisected at the point (2,1)(2,1)
View answer and explanation
Correct option: C
5x+4y=135 x+4 y=13
Option C (5x+4y=135 x+4 y=13) is correct because it is obtained by using the equation T=S1T=S_1 for the chord bisected at a given point.
193 Easy
193 of 392
Application +4 | -1
If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latusrectum is
View answer and explanation
Correct option: B
323 \sqrt{2}
Option B (323 \sqrt{2}) is correct because it is the length of the latus rectum calculated from the given distance between foci and directrices.
194 Easy
194 of 392
Application +4 | -1
The equation of normal at the point (0,3)(0,3) of the ellipse 9x2+5y2=459 x^2+5 y^2=45 is
View answer and explanation
Correct option: D
yy-axis
Option D (yy-axis) is correct because the normal to the ellipse at the point (0,3) is along the y-axis.
195 Easy
195 of 392
Application +4 | -1
If the distance between the directrices be thrice the distance between the foci, then the eccentricity of ellipse is
View answer and explanation
Correct option: C
1/31 / \sqrt{3}
Option C (-4) is correct because it is found using the condition c=−am2c = -a m^2 for tangency.
196 Easy
196 of 392
Application +4 | -1
If the coordinates of two points A   and  BA\,\,\,\mathrm{and }\,\,B are (7,0)\left( {\sqrt {7,} 0} \right) and (−7,0)\left( { - \sqrt {7,}0 } \right) respectively and PP is any point on the conic, 9x2+16y2=144,9{x^2} + 16{y^2} = 144, then PA+PBPA + PB is equal to
View answer and explanation
Correct option: B
8
The correct answer is B (8) because it represents the sum of the distances from any point P on the ellipse to the two foci A and B, which is equal to 2a, where a is the semi-major axis.
197 Easy
197 of 392
Application +4 | -1
Let S(3,4)S(3,4) and S′(9,12)S^{\prime}(9,12) be two foci of an ellipse. If the coordinates of the foot of the perpendicular from focus SS to a tangent to the ellipse is (1,−4)(1,-4), then the eccentricity of the ellipse, is
View answer and explanation
Correct option: D
5/135 / 13
Option D (5/135 / 13) is correct because the eccentricity is calculated using the distance between foci and the property that the foot of the perpendicular from the focus to a tangent lies on the auxiliary circle.
198 Easy
198 of 392
Application +4 | -1
The position of the point (1,3)(1,3) w.r.t. the ellipse 4x2+9y2−16x4 x^2+9 y^2-16 x −54y+61=0-54 y+61=0 is
View answer and explanation
Correct option: C
On the major axis
Option C (On the major axis) is correct because substituting the point (1,3) into the ellipse equation results in a negative value, indicating the point lies inside the ellipse, specifically on the major axis.
199 Easy
199 of 392
Application +4 | -1
Eccentricity of the ellipse x2+2y2−2x+3y+2=0{x^2} + 2{y^2} - 2x + 3y + 2 = 0 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 of the ellipse is found by rewriting the equation in standard form and using the formula e=1−b2a2e = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}}.
200 Easy
200 of 392
Application +4 | -1
If the line y=mx+cy=m x+c touches the ellipse x2b2+y2a2=1\frac{x^2}{b^2}+\frac{y^2}{a^2}=1, then find the value of cc.
View answer and explanation
Correct option: A
±b2m2+a2\pm \sqrt{b^2 m^2+a^2}
Option A (±b2m2+a2\pm \sqrt{{b^2 m^2+a^2}}) is correct because, for the ellipse x2b2+y2a2=1\frac{{x^2}}{{b^2}} + \frac{{y^2}}{{a^2}} = 1, the condition for tangency of the line y = mx + c is c=±a2+b2m2c = \pm \sqrt{{a^2 + b^2 m^2}} due to the interchange of a and b.
201 Easy
201 of 392
Application +4 | -1
Equation of the tangents to the ellipse x29+y216=1\frac{x^2}{9}+\frac{y^2}{16}=1 which are parallel to the line x+y+1=0x+y+1=0, is
View answer and explanation
Correct option: A
x+y+5=0x+y+5=0
Option A (x+y+5=0x+y+5=0) is correct because it satisfies the condition for tangency to the ellipse and is parallel to the given line.
202 Easy
202 of 392
Application +4 | -1
The curve represented by x=2(cos⁡t+sin⁡t)x=2(\cos t+\sin t), y=5(cos⁡t−sin⁡t)y=5(\cos t-\sin t), is
View answer and explanation
Correct option: C
an ellipse
Option C (an ellipse) is correct because the given parametric equations can be manipulated to form the equation of an ellipse.
203 Easy
203 of 392
Analyzing +4 | -1
Let SS and S′S^{\prime} be two foci of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1. If a circle described on SS′SS' as diameter intersects the ellipse in real and distinct points, then the eccentricity ee of the ellipse satisfies
View answer and explanation
Correct option: B
e∈(1/2,1)e \in(1 / \sqrt{2}, 1)
Option A (tan⁡−1(a−b2ab)\tan ^{-1}\left(\frac{a-b}{2 \sqrt{a b}}\right)) is correct because it represents the greatest acute angle between the tangents to the ellipse and the circle at the points where a common ordinate cuts them.
204 Easy
204 of 392
Application +4 | -1
The distances from the foci of point P(x1,y1)P\left( {{x_1},{y_1}} \right) on the ellipse x29+y225=1\frac{{{x^2}}}{9} + \frac{{{y^2}}}{{25}} = 1 are
View answer and explanation
Correct option: C
5±45y15 \pm \frac{4}{5}{y_1}
Option C (5±45y15 \pm \frac{4}{5}{y_1}) is correct because it represents the focal distances of a point on the ellipse, considering that the major axis is along the y-axis.
205 Easy
205 of 392
Application +4 | -1
If two tangents drawn to the ellipse x2+4y2=4x^2+4 y^2=4 intersect perpendicularly at PP, then the locus of PP is
View answer and explanation
Correct option: B
x2+y2=5x^2+y^2=5
Option B (x2+y2=5x^2+y^2=5) is correct because the locus of P is the director circle of the ellipse, and its equation is x2+y2=a2+b2x^2 + y^2 = a^2 + b^2.
206 Easy
206 of 392
Application +4 | -1
Let PP be a point in the first quadrant lying on the ellipse 9x2+16y2=1449 x^2+16 y^2=144, such that the tangent at PP to the ellipse is inclined at an angle of 135∘135^{\circ} to the positive direction of xx-axis. Then the coordinates of PP are
View answer and explanation
Correct option: A
(165,95)\left(\frac{16}{5}, \frac{9}{5}\right)
Option A ((16/5, 9/5)) is correct because it satisfies the ellipse equation and the tangent at that point has a slope of -1.
207 Easy
207 of 392
Application +4 | -1
The angle between the pair of tangents drawn to the ellipse 3x2+2y2=53{x^2} + 2{y^2} = 5 from the point (1,2)\left( {1,2} \right) is
View answer and explanation
Correct option: C
tan⁡−1(125){\tan ^{ - 1}}\left( {\frac{{12}}{{\sqrt 5 }}} \right)
Option C (tan⁡−1(125){\tan ^{ - 1}}(\frac{{12}}{{\sqrt 5 }})) is correct because it is the angle between the pair of tangents calculated using the joint equation of tangents.
208 Easy
208 of 392
Application +4 | -1
The distance of the point (6cos⁡θ,2sin⁡θ)(\sqrt{6} \cos \theta, \sqrt{2} \sin \theta) on the ellipse from the centre of the ellipse is 2 if θ=\theta=
View answer and explanation
Correct option: B
π/4\pi/4
Option B (π/4\pi/4) is correct because substituting the point into the distance formula and solving for θ\theta yields π/4\pi/4.
209 Easy
209 of 392
Application +4 | -1
If the coordinates of the centre, a focus and adjacent vertex are (2,−3),(3,−3)\left( {2, - 3} \right),\left( {3, - 3} \right) and (4,−3)\left( {4, - 3} \right) respectively, then the equation of the ellipse is
View answer and explanation
Correct option: A
(x−2)24+(y+3)23=1\frac{{{{\left( {x - 2} \right)}^2}}}{4} + \frac{{{{\left( {y + 3} \right)}^2}}}{3} = 1
Option A ((x−2)24+(y+3)23=1\frac{{{{\left( {x - 2} \right)}^2}}}{4} + \frac{{{{\left( {y + 3} \right)}^2}}}{3} = 1) is correct because it correctly represents an ellipse with the given center, focus, and vertex.
210 Easy
210 of 392
Analyzing +4 | -1
x−2y+4=0x-2 y+4=0 is a common tangent to y2=4xy^2=4 x and x24+y2b2=1\frac{x^2}{4}+\frac{y^2}{b^2}=1. Then, the value of bb and the other common tangent are given by
View answer and explanation
Correct option: A
b=3;x+2y+4=0b=\sqrt{3} ; x+2 y+4=0
Option A (b=3;x+2y+4=0b=\sqrt{{3}} ; x+2 y+4=0) is correct because the given line is tangent to both the parabola and the ellipse, and the other common tangent is found using symmetry.
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