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1
JEE Main 2021 (Online) 16th March Morning Shift
+4
-1
If the three normals drawn to the parabola, y2 = 2x pass through the point (a, 0) a $$\ne$$ 0, then 'a' must be greater than :
A
$${1 \over 2}$$
B
1
C
$$-$$1
D
$$-$$$${1 \over 2}$$
2
JEE Main 2021 (Online) 16th March Morning Shift
+4
-1
The locus of the midpoints of the chord of the circle, x2 + y2 = 25 which is tangent to the hyperbola, $${{{x^2}} \over 9} - {{{y^2}} \over {16}} = 1$$ is :
A
(x2 + y2)2 $$-$$ 9x2 + 16y2 = 0
B
(x2 + y2)2 $$-$$ 9x2 + 144y2 = 0
C
(x2 + y2)2 $$-$$ 16x2 + 9y2 = 0
D
(x2 + y2)2 $$-$$ 9x2 $$-$$ 16y2 = 0
3
JEE Main 2021 (Online) 25th February Evening Shift
+4
-1
The shortest distance between the line x $$-$$ y = 1 and the curve x2 = 2y is :
A
0
B
$${1 \over 2{\sqrt 2 }}$$
C
$${1 \over {\sqrt 2 }}$$
D
$${1 \over 2}$$
4
JEE Main 2021 (Online) 25th February Evening Shift
A hyperbola passes through the foci of the ellipse $${{{x^2}} \over {25}} + {{{y^2}} \over {16}} = 1$$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is :
$${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$
$${{{x^2}} \over 9} - {{{y^2}} \over 16} = 1$$
$${{{x^2}} \over 9} - {{{y^2}} \over 25} = 1$$
x2 $$-$$ y2 = 9