Implicit Differentiation - Exercise 3

Exercise 3. Show that if a normal line to each point on an ellipse passes through the center of an ellipse, then the ellipse is a circle.

Answer. An equation of the ellipse is given by

\begin{displaymath}\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\end{displaymath}

where we assumed that (0,0) is the center. We may always do that. Let (x0,y0) be a point on the ellipse. The slope of the tangent line to the ellipse at this point will be obtained through implicit differentiation. Indeed, we have

\begin{displaymath}2 \frac{x}{a^2} + \frac{2 y y'}{b^2} = 0\end{displaymath}

or equivalently

\begin{displaymath}y' = - \frac{x/a^2}{y/b^2} \;\cdot\end{displaymath}

So the slope of the tangent line at the point (x0,y0) is $\displaystyle - \frac{x_0/a^2}{y_0/b^2}$ which gives the slope of the normal line as $\displaystyle \frac{y_0/b^2}{x_0/a^2}$. Hence the equation of the normal line is

\begin{displaymath}y - y_0 = \frac{y_0/b^2}{x_0/a^2} (x-x_0)\end{displaymath}

or equivalently

\begin{displaymath}\frac{x_0}{a^2} (y - y_0) = \frac{y_0}{b^2} (x-x_0)\;.\end{displaymath}

Assuming that (0,0) is on any normal line, we get

\begin{displaymath}- \frac{x_0}{a^2} y_0 = - \frac{y_0}{b^2} x_0\;.\end{displaymath}

If we choose a point (x0,y0) such that $x_0 y_0 \neq 0$, we will get

a2 = b2

which clearly implies that the ellipse is a circle.

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