1
JEE Main 2022 (Online) 25th June Evening Shift
+4
-1

The line y = x + 1 meets the ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$$ at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)2 is equal to :

A
20
B
12
C
11
D
8
2
JEE Main 2022 (Online) 24th June Evening Shift
+4
-1

Let the maximum area of the triangle that can be inscribed in the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 4} = 1,\,a > 2$$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be $$6\sqrt 3$$. Then the eccentricity of the ellipse is :

A
$${{\sqrt 3 } \over 2}$$
B
$${1 \over 2}$$
C
$${1 \over {\sqrt 2 }}$$
D
$${{\sqrt 3 } \over 4}$$
3
JEE Main 2021 (Online) 1st September Evening Shift
+4
-1
Out of Syllabus
Let $$\theta$$ be the acute angle between the tangents to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 1} = 1$$ and the circle $${x^2} + {y^2} = 3$$ at their point of intersection in the first quadrant. Then tan$$\theta$$ is equal to :
A
$${5 \over {2\sqrt 3 }}$$
B
$${2 \over {\sqrt 3 }}$$
C
$${4 \over {\sqrt 3 }}$$
D
2
4
JEE Main 2021 (Online) 31st August Evening Shift
+4
-1
The locus of mid-points of the line segments joining ($$-$$3, $$-$$5) and the points on the ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$$ is :
A
$$9{x^2} + 4{y^2} + 18x + 8y + 145 = 0$$
B
$$36{x^2} + 16{y^2} + 90x + 56y + 145 = 0$$
C
$$36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$$
D
$$36{x^2} + 16{y^2} + 72x + 32y + 145 = 0$$
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