Question

Find the point on the parabola $y^2=2x$ that is closest to the point $(1,4)$.

Answer 1

Closest point on the parabola to the point will be on the normal from the parabola.

Consider $y^2=2x$

Differentiate with respect to $x$

$\Rightarrow 2y\frac{dy}{dx}=2$

$\Rightarrow \frac{dy}{dx}=\frac{1}{y}$

Slope of normal $=-y$

Equation of normal through $(1,4)$ is $y-4=-y(x-1)$

At intersection $x=\frac{y^2}{2}$

$\Rightarrow y-4=-y(\frac{y^2}{2}-1)$

$\Rightarrow y-4=-\frac{y^3}{2}+y$

$\Rightarrow y^3=8$

$\Rightarrow y=2$
Since $x=\frac{y^2}{2}$ and $y=2$ we get $x=2$

Hence the closest point on the parabola is $(2,2)$