Find the point on the curve y^2=8 x for which the abscissa and ordinate change at the same rate.

Here, $\displaystyle \frac{d \mathrm{x}}{d t}=\frac{d \mathrm{y}}{d t}$ Given $\displaystyle \mathrm{y}^{2}=8 \mathrm{x}$ gives $\displaystyle 2 \mathrm{y} \cdot \frac{d \mathrm{y}}{d t}=8 \frac{d \mathrm{x}}{d t}$ $$\Rightarrow 2 \mathrm{y}=8 \text { or } \mathrm{y}=4 $$ Also, $\displaystyle \mathrm{y}=4$ gives $\displaystyle \mathrm{x}=2$. Thus, the point $\displaystyle (2,4)$