JEE Main 2022 (Online) 24th June Evening Shift | Ellipse Question 64 | Mathematics

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 :

$${{\sqrt 3 } \over 2}$$

$${1 \over 2}$$

$${1 \over {\sqrt 2 }}$$

$${{\sqrt 3 } \over 4}$$

JEE Main 2021 (Online) 1st September Evening Shift

MCQ (Single Correct Answer)

-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 :

$${5 \over {2\sqrt 3 }}$$

$${2 \over {\sqrt 3 }}$$

$${4 \over {\sqrt 3 }}$$

JEE Main 2021 (Online) 31st August Evening Shift

MCQ (Single Correct Answer)

-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 :

$$9{x^2} + 4{y^2} + 18x + 8y + 145 = 0$$

$$36{x^2} + 16{y^2} + 90x + 56y + 145 = 0$$

$$36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$$

$$36{x^2} + 16{y^2} + 72x + 32y + 145 = 0$$

JEE Main 2021 (Online) 31st August Evening Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

An angle of intersection of the curves, $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ and x² + y² = ab, a > b, is :

$${\tan ^{ - 1}}\left( {{{a + b} \over {\sqrt {ab} }}} \right)$$

$${\tan ^{ - 1}}\left( {{{a - b} \over {2\sqrt {ab} }}} \right)$$

$${\tan ^{ - 1}}\left( {{{a - b} \over {\sqrt {ab} }}} \right)$$