Question

Find the condition that the line x/m+y/n=1 is tangent to the ellipse x^2/a^2+y^2/b^2=1

Answer 1

Any line tangent to the ellipse at \((x_0,y_0)\) will have slope \(\dfrac{dy}{dx}\) evaluated at \((x_0,y_0)\), which you can find by computing the derivative for the ellipse equation:
\[\begin{align*}\frac{d}{dx}\left[\frac{x^2}{a^2}+\frac{y^2}{b^2}\right]&=\frac{d}{dx}[1]\\[1ex]
\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}&=0\\[1ex]
\frac{dy}{dx}&=-\frac{b^2x}{a^2y}\end{align*}\]The line \(\dfrac{x}{m}+\dfrac{y}{n}=1\) has slope \(-\dfrac{n}{m}\), so at the point \((x_0,y_0)\), the tangent line satisfies
\[-\frac{n}{m}=-\frac{b^2x_0}{a^2y_0}\]