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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 121-150 of 392 questions

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121 Easy
121 of 392
Analyzing +4 | -1
Ellipses which are drawn with the same two perpendicular lines as axes and with the sum of the reciprocals of squares of the lengths of their semi-major axis and semi-minor axis equal to a constant have only
View answer and explanation
Correct option: B
four points in common
Option C (-a4b4\frac{a^4}{b^4}) is correct because the condition that the chords of contact are at right angles leads to this relationship between the coordinates of the points and the ellipse parameters.
122 Easy
122 of 392
Application +4 | -1
The locus of a point whose distance from the point (3,0)(3,0) is 3/53 / 5 times its distance from the line x=px=p is an ellipse with centre at the origin. The value of pp is
View answer and explanation
Correct option: C
253\frac{25}{3}
Option C (253\frac{25}{3}) is correct because it represents the x-coordinate of the directrix of the ellipse.
123 Easy
123 of 392
Understanding +4 | -1
The equation (x−2)2+y2+(x+2)2+y2=4\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4 represents
View answer and explanation
Correct option: A
line segment
Option A (line segment) is correct because the equation represents the sum of distances from two fixed points being constant, which defines a line segment between those points.
124 Easy
124 of 392
Application +4 | -1
If e1e_1 is the eccentricity of the conic 9x2+4y2=369 x^2+4 y^2=36 and e2e_2 is the eccentricity of the conic 9x2−4y29 x^2-4 y^2 =36=36, then -
View answer and explanation
Correct option: B
e12+e22>3e_1^2+e_2^2>3
Option B (e12+e22>3e_1^2+e_2^2>3) is correct because the sum of the squares of the eccentricities of the ellipse and hyperbola is greater than 3.
125 Easy
125 of 392
Remembering +4 | -1
In the standard form of an ellipse sum of the focal distances of a point is
View answer and explanation
Correct option: C
2a2a
Option C (2a2a) is correct because the sum of the focal distances of any point on an ellipse is equal to the length of the major axis, which is 2a.
126 Easy
126 of 392
Analyzing +4 | -1
If pp and p′p^{\prime} denote the lengths of the perpendicular from a focus and the centre of an ellipse with semi-major axis of length aa, respectively, on a tangent to the ellipsic and rr denotes the focal distance of the point, then
View answer and explanation
Correct option: A
ap=rp′a p=r p^{\prime}
Option A (±1\pm 1) is correct because the common chord of the two parabolas is given by x2−y2=0x^2 - y^2 = 0, which gives the lines y=±xy = \pm x, hence the slope is ±1\pm 1.
127 Easy
127 of 392
Application +4 | -1
The distance between the foci of the ellipse 5x2+9y2=45,5{x^2} + 9{y^2} = 45, is
View answer and explanation
Correct option: B
4
Option B (4) is correct because the distance between the foci is calculated using the eccentricity and the major axis length.
128 Easy
128 of 392
Analyzing +4 | -1
Let d1d_1 and d2d_2 be the lengths of perpendiculars drawn from foci S and S′S^{\prime} of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 to the tangent at any point PP on the ellipse. Then, SP:SP′=SP : SP' =
View answer and explanation
Correct option: A
d1:d2d_1: d_2
Option B (e∈(1/2,1)e \in(1 / \sqrt{2}, 1)) is correct because the eccentricity e of the ellipse must lie in the interval (1/2,1)(1 / \sqrt{2}, 1) for the circle to intersect the ellipse in real and distinct points.
129 Easy
129 of 392
Application +4 | -1
The coordinates of the foci of the ellipse 4x2+9y2=14x^2 + 9y^2 = 1 are
View answer and explanation
Correct option: B
(±5/6,0)\left( { \pm \sqrt 5 /6,0} \right)
Option B ((±5/6,0)\left( { \pm \sqrt 5 /6,0} \right)) is correct because the foci coordinates are calculated using the formula (±ae,0\pm ae, 0) where a and e are derived from the ellipse equation.
130 Easy
130 of 392
Application +4 | -1
The equation x22−r+y2r−5+1=0\frac{x^2}{2-r}+\frac{y^2}{r-5}+1=0 represents an ellipse if
View answer and explanation
Correct option: B
2<r<52 < r < 5
Option B (2<r<52 < r < 5) is correct because for the equation to represent an ellipse, both denominators must be positive.
131 Easy
131 of 392
Understanding +4 | -1
If ax2+by2+2gx+2fy+c=0a{x^2} + b{y^2} + 2\mathrm{g}x + 2fy + c = 0 represents an ellipse, then
View answer and explanation
Correct option: D
It’s axes are parallel to coordinates axes
The correct answer is D (It’s axes are parallel to coordinates axes) because the absence of an xyxy term allows for completing the square in both xx and yy, indicating axes parallel to the coordinate axes.
132 Easy
132 of 392
Analyzing +4 | -1
If polar of x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is always touching the parabola y2=4axy^2=4 a x, then locus of the pole is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the correct locus equation is a3y2=−b4xa^3 y^2 = -b^4 x, which does not match any of the given options. The locus is found by applying the tangency condition.
133 Easy
133 of 392
Application +4 | -1
The locus of the point of intersection of mutually perpendicular tangent 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
A circle
Option C (A circle) is correct because the locus of the point of intersection of mutually perpendicular tangents to the ellipse is the director circle, which is a circle.
134 Easy
134 of 392
Application +4 | -1
The values of λ\lambda for which the line y=x+λy=x+\lambda touches the ellipse 9x2+16y2=1449 x^2+16 y^2=144, are
View answer and explanation
Correct option: A
±5\pm 5
Option A (±5\pm 5) is correct because these are the values of lambda that satisfy the tangency condition for the line y = x + lambda to the given ellipse.
135 Easy
135 of 392
Application +4 | -1
The equation of the chord of contact of the pair of tangents drawn to the ellipse 4x2+9y2=364 x^2+9 y^2=36 from the point (m,n)(m, n), where m⋅n=m+n,m,nm \cdot n=m+n, m, n being non-zero positive integers is
View answer and explanation
Correct option: C
4x+9y=184 x+9 y=18
Option C (4x+9y=184 x+9 y=18) is correct because substituting m=2m=2 and n=2n=2 into the chord of contact equation yields the given result.
136 Easy
136 of 392
Application +4 | -1
If y=mx+4y=m x+4 is a tangent to the ellipse x225+y24=1\frac{x^2}{25}+\frac{y^2}{4}=1, then 625m4+25m2−156625 m^4+25 m^2-156 is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because substituting the condition for tangency into the given expression results in zero.
137 Easy
137 of 392
Application +4 | -1
If the tangents are drawn to the ellipse x2+2y2=2x^2+2 y^2=2, then the locus of the mid-point of the intercept made by the tangents between the coordinate axes, is
View answer and explanation
Correct option: A
12x2+14y2=1\frac{1}{2 x^2}+\frac{1}{4 y^2}=1
Option A (12x2+14y2=1\frac{{1}}{{2 x^2}}+\frac{{1}}{{4 y^2}}=1) is correct because it represents the locus of the midpoint of the intercept made by the tangents between the coordinate axes.
138 Easy
138 of 392
Application +4 | -1
Eccentricity of hyperbola x2k+y2k2=1 (k<0)\frac{{{x^2}}}{k} + \frac{{{y^2}}}{{{k^2}}} = 1\,(k < 0) is
View answer and explanation
Correct option: D
1−1k\sqrt {1 - \frac{1}{k}}
Option D (1−1k\sqrt {1 - \frac{1}{k}}) is correct because it represents the eccentricity of the hyperbola given the condition k < 0.
139 Easy
139 of 392
Application +4 | -1
Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, x24+y22=1\frac{{{x^2}}}{4} + \frac{{{y^2}}}{2} = 1 from any of its foci?
View answer and explanation
Correct option: C
(−1,3)( - 1,\sqrt 3 )
Option C ((−1,3)( - 1,\sqrt 3 )) is correct because it satisfies the equation of the auxiliary circle of the given ellipse, which is the locus of the foot of the perpendicular from the foci to any tangent.
140 Easy
140 of 392
Analyzing +4 | -1
If the normals at P(x1,y1),Q(x2,y2)P\left(x_1, y_1\right), Q\left(x_2, y_2\right) and R(x3,y3)R\left(x_3, y_3\right) to the ellipse are concurrent, then
View answer and explanation
Correct option: B
∣x1y1x1y1x2y2x2y2x3y3x3y3∣=0\left|\begin{array}{lll}x_1 & y_1 & x_1 y_1 \\ x_2 & y_2 & x_2 y_2 \\ x_3 & y_3 & x_3 y_3\end{array}\right|=0
Option B (∣x1y1x1y1x2y2x2y2x3y3x3y3∣=0\left|\begin{array}{lll}x_1 & y_1 & x_1 y_1 \\ x_2 & y_2 & x_2 y_2 \\ x_3 & y_3 & x_3 y_3\end{array}\right|=0) is correct because it represents the condition for concurrency of normals at three points on the ellipse.
141 Easy
141 of 392
Understanding +4 | -1
If CPC P and CDCD are semi-conjugate diameters of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, then CP2+CD2=C P^2+C D^2=
View answer and explanation
Correct option: B
a2+b2a^2+b^2
Option B (a2+b2a^2+b^2) is correct because the sum of the squares of semi-conjugate diameters of an ellipse is constant and equal to a2+b2a^2 + b^2.
142 Easy
142 of 392
Application +4 | -1
In the ellipse, the minor axis is 8 and eccentricity is 5/3\sqrt{5} / 3. Then major axis is
View answer and explanation
Correct option: B
12
Option B (12) is correct because the major axis is calculated using the eccentricity and minor axis of the ellipse.
143 Easy
143 of 392
Application +4 | -1
If α\alpha and β\beta are eccentric angles of the ends of aa focal chord of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, then tan⁡α2tan⁡β2\tan \frac{\alpha}{2} \tan \frac{\beta}{2} is equal to
View answer and explanation
Correct option: B
e−1e+1\frac{e-1}{e+1}
Option B (e−1e+1\frac{e-1}{e+1}) is correct because it represents the relationship between the tangents of half angles and the eccentricity.
144 Easy
144 of 392
Application +4 | -1
The normal at the point (3,23)(3,2 \sqrt{3}) on the ellipse x236+y216=1\frac{x^2}{36}+\frac{y^2}{16}=1 meets the major axis of the ellipse at (α,0)(\alpha, 0), the value of α\alpha is:
View answer and explanation
Correct option: A
53\frac{5}{3}
Option A (53\frac{5}{3}) is correct because substituting y=0 into the equation of the normal to the ellipse at the given point yields x = 5/3.
145 Easy
145 of 392
Application +4 | -1
In an ellipse length of minor axis is 8 and eccentricity is 53\frac{{\sqrt 5 }}{3}. The length of major axis is
View answer and explanation
Correct option: B
12
Option B (12) is correct because the length of the major axis is calculated using the relationship between the minor axis, eccentricity, and major axis of an ellipse.
146 Easy
146 of 392
Application +4 | -1
Locus of the point which divides double ordinate of the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 in the ratio 1:21:2 internally, is
View answer and explanation
Correct option: B
x2a2+9y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{9{y^2}}}{{{b^2}}} = 1
Option B (x2a2+9y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{9{y^2}}}{{{b^2}}} = 1) is correct because it represents the locus of the point dividing the double ordinate in the given ratio.
147 Easy
147 of 392
Application +4 | -1
The vertices of the ellipse 9x2+4y2−18x−27=09 x^2+4 y^2-18 x-27=0, are
View answer and explanation
Correct option: B
(1,±3)(1, \pm 3)
Option B ((1,±3)(1, \pm 3)) is correct because it represents the vertices of the ellipse after shifting the origin.
148 Easy
148 of 392
Application +4 | -1
A focus of an ellipse is at the origin. The directrix is the line x=4x = 4 and the eccentricity is 12\frac{1}{2}, then length of semi major axis is
View answer and explanation
Correct option: B
83\frac{8}{3}
Option B (83\frac{8}{3}) is correct because it is calculated using the relationship between the distance from the center to the directrix and the eccentricity of the ellipse.
149 Easy
149 of 392
Application +4 | -1
All ellipse has its centre at (1,−1)\left( {1, - 1} \right) and semi-major axis =8= 8 and it passes through the point (1,3)\left( {1,3} \right). The equation of the ellipse is
View answer and explanation
Correct option: B
(x−1)264+(y+1)216=1\frac{{{{\left( {x - 1} \right)}^2}}}{{64}} + \frac{{{{\left( {y + 1} \right)}^2}}}{{16}} = 1
Option B ((x−1)264+(y+1)216=1\frac{{{{\left( {x - 1} \right)}^2}}}{{64}} + \frac{{{{\left( {y + 1} \right)}^2}}}{{16}} = 1) is correct because it satisfies the given conditions: center at (1, -1), semi-major axis of 8, and passing through the point (1, 3).
150 Easy
150 of 392
Application +4 | -1
The equation of the ellipse whose foci are (± 2,0)\left( { \pm \,2,0} \right) and eccentricity 12\frac{1}{2} is
View answer and explanation
Correct option: B
x216+y212=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{12}} = 1
Option B (x216+y212=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{12}} = 1) is correct because it satisfies the conditions for the ellipse with given foci and eccentricity.
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