JEE Advance - Mathematics (1987 - No. 24)
Find the point on the curve $$\,\,\,4{x^2} + {a^2}{y^2} = 4{a^2},\,\,\,4 < {a^2} < 8$$
that is farthest from the point $$(0, -2)$$.
that is farthest from the point $$(0, -2)$$.
(0, 2)
(a, 0)
(0, -2)
(2, 0)
(a/2, a/2)