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Q.
The point on the curve $x^2=2 y$ which is nearest to the point $(0,5)$ is
Application of Derivatives
Solution:
Let $d$ be the distance of the point $(x, y)$ on $x^2=2 y$ from the point $(0,5)$, then
$d =\sqrt{(x-0)^2+(y-5)^2} =\sqrt{2 y+(y-5)^2}$.......(i)
$ = \sqrt{2y + (y - 5)^2} $ (putting $x^2 = 2y)$
$ =\sqrt{y^2-8 y+25} $
$ =\sqrt{y^2-8 y+4^2+9} $
$=\sqrt{(y-4)^2+9}$
$d$ is least when $(y-4)^2=0$ i.e., when $y=4$
When $y=4$, then $x^2=2 \times 4$
$\Rightarrow x=\pm \sqrt{8}=\pm 2 \sqrt{2}$
$\therefore$ The points $(2 \sqrt{2}, 4)$ and $(-2 \sqrt{2}, 4)$ on the given curve are nearest to the point $(0,5)$. So, (a) is the correct option.