Question

A particle moves along the curve $x^2=2y$. At what point, ordinate increases at the same rate as abscissa increases ?

Answer 1

$x^2=2y$ .......(1)

$2x\frac{dx}{dt}=2\frac{dy}{dt}$

$2x\frac{dx}{dt}=2\frac{dx}{dt}$ $[\frac{dx}{dt}=\frac{dy}{dt}]$

$x=1$

From 1, $y=\frac{1}{2}$

So, point is $(1,\frac{1}{2})$.