The point on the curve x^2=2 y in the I quadrant which is nearest to…

The correct answer is Option (3) → $(2 \sqrt{2}, 4)$

let point on curve be (x, y)

so distance

$d(x)=\sqrt{(x-0)^2+(y-5)^2}$

as from curve $y=\frac{x^2}{2}$

so $d(x)=\sqrt{x^2+(\frac{x^2}{2}-5)^2}$

we need to minimise d(x)

so minimising $f(x)=x^2+(\frac{x^2}{2}-5)^2$ also works

$f'(x)=2x+2(\frac{x^2}{2}-5)×\frac{2x}{2}=0$

$f'(x)=2x+x^3-10x=0$ or $x^3=8x$

so $x=0,2\sqrt{2}$

$f''(x)=3x^2-8$

$f''(0)=-8<0$ (point of maxima)

$f''(2\sqrt{2})=16>0$ (point of minima)

at $x=2\sqrt{2},y=4$

point nearest to (0, 5) is $(2 \sqrt{2}, 4)$