The point on the parabola $y^{2}=64 x$ which is nearest to the line $4 x+3 y+35=0$ has coordinates

Given equation of parabola is $$ y^{2}=64 x $$ The point at which the tangent to the curve is parallel to the line is the nearest point on the curve. On differentiating both sides of Eq. (i), we get $2 y \frac{d y}{d x}=64$ $\Rightarrow \quad \frac{d y}{d x}=\frac{32}{y}$ Also, slope of the given line is $-\frac{4}{3}$. $\therefore$ $$ -\frac{4}{3}=\frac{32}{y} \Rightarrow y=-24 $$ From Eq. (i), $(-24)^{2}=64 x \Rightarrow x=9$ $\therefore$ Hence, the required point is (9,-24)