1
JEE Main 2021 (Online) 1st September Evening Shift
+4
-1
Consider the parabola with vertex $$\left( {{1 \over 2},{3 \over 4}} \right)$$ and the directrix $$y = {1 \over 2}$$. Let P be the point where the parabola meets the line $$x = - {1 \over 2}$$. If the normal to the parabola at P intersects the parabola again at the point Q, then (PQ)2 is equal to :
A
$${{75} \over 8}$$
B
$${{125} \over {16}}$$
C
$${{25} \over 2}$$
D
$${{15} \over 2}$$
2
JEE Main 2021 (Online) 1st September Evening Shift
+4
-1
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
3
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
$$36{x^2} + 16{y^2} + 72x + 32y + 145 = 0$$
4
JEE Main 2021 (Online) 31st August Evening Shift
+4
-1
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 :
A
$${\tan ^{ - 1}}\left( {{{a + b} \over {\sqrt {ab} }}} \right)$$
B
$${\tan ^{ - 1}}\left( {{{a - b} \over {2\sqrt {ab} }}} \right)$$
C
$${\tan ^{ - 1}}\left( {{{a - b} \over {\sqrt {ab} }}} \right)$$
D
$${\tan ^{ - 1}}\left( {2\sqrt {ab} } \right)$$
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