Problem source

The normal to the curve $y=\mathrm{f}(x)$ at the point $P$ with coordinates $(x,\mathrm{f}(x))$ cuts the $y$-axis at the point $Q$.
Derive an expression in terms of $x$, $\mathrm{f}(x)$ and $\mathrm{f}'(x)$ for the $y$-coordinate of $Q$. 
If, for all $x$, $PQ=\sqrt{\mathrm{e}^{x^{2}}+x^{2}}$, find a differential equation satisfied by $\mathrm{f}(x)$. If the curve also has a minimum point $(0,-2)$, find its equation.

Solution source

The normal to the curve $y = f(x)$ has gradient $-\frac{1}{f'(x)}$ and so has equation:

\begin{align*}
&& \frac{Y - f(x)}{X - x} &= -\frac{1}{f'(x)} \\
\Rightarrow && Y &= -\frac{1}{f'(x)}X + \frac{x}{f'(x)}+f(x)
\end{align*}

Hence the $Q$ is $\displaystyle \left (0, f(x) + \frac{x}{f'(x)} \right)$. 

\begin{align*}
&& |PQ|^2 &= x^2 + \frac{x^2}{(f'(x))^2} \\
\Rightarrow && x^2 + e^{x^2} &= x^2 + \frac{x^2}{(f'(x))^2} \\
\Rightarrow && (f'(x))^2 &=x^2 e^{-x^2}
\end{align*}

Therefore $f'(x) = \pm x e^{-x^2/2}$. If $f(x)$ has a minimum at $(0,-2)$ then $f''(0) > 0$, and $f''(x) = \pm (e^{-x^2/2} - x^2e^{-x^2/2}) = \pm e^{-x^2/2}(1-x^2)$ so we should take the positive branch of the solution, ie $f'(x) = xe^{-x^2/2}$.

Therefore $f(x) = - e^{-x^2/2}+C$. Since $f(0) = -2$ we must have $-2 = -1 + C$, ie $C = -1$.

Therefore $f(x) = -1 - e^{-x^2/2}$