The points on the curve 2x=y^2 which are nearest to the point (5, 0)…

The correct answer is Option (3) → A, D only

Let point on curve be (x, y)

needs to be minimum from curve $y^2=2x$

so $d=(x-5)^2+y^2$

$d=(x-5)^2+2x$

$d'=2(x-5)+2=0$

so $x=4$

so $y^2=2×4=8$

$y=±2\sqrt{2}$

Points $(4, ±2\sqrt{2})$ lie in Ist and IVth quadrant (A, D)