Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 50

a.) Where does the normal line to the ellipse $x^2 - xy + y^2 = 3$ at the point $(-1,1)$ intersect the ellipse a second time?

Taking the derivative of the curve implicitly we have,


$\begin{equation} \begin{aligned} & \frac{d}{dx} (x^2) - \frac{d}{dx} (xy) + \frac{d}{dx} (y^2) = \frac{d}{dx} (3) \\ \\ & 2x - \left[ \frac{d}{dx} (x) \cdot y + x \cdot \frac{d}{dx} (y) \right] + \frac{d}{dy} (y^2) \frac{dy}{dx} = 0 \\ \\ & 2x - \left[ y + x \frac{dy}{dx} \right] + 2y \frac{dy}{dx} = 0 \\ \\ & 2x - y - x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0 \\ \\ & \frac{dy}{dx} = \frac{2x - y}{x - 2y} \end{aligned} \end{equation}$


Recall that the slope of the normal line is equal to the negative reciprocal of the tangent line so..

$\displaystyle y'_N = \frac{- (x - 2y)}{2x - y} = \frac{-x + 2y}{2x - y}$

@ Point $(-1, 1)$


$\begin{equation} \begin{aligned} y'_N =& \frac{-(-1) + 2 (1)}{2(-1) - (1)} \\ \\ y'_N =& -1 \end{aligned} \end{equation}$


Thus the equation of the normal line is..


$\begin{equation} \begin{aligned} y - (1) =& -1 (x - (-1)) \\ \\ y - \cancel{1} =& -x - \cancel{1} \\ \\ y =& -x \end{aligned} \end{equation}$


Now, since the normal line and the curve intersects, we can substitute $y$ at the equation of the ellipse, so...


$\begin{equation} \begin{aligned} x^2 - xy + y^2 =&3 \\ \\ x^2 - x(-x) + (-x)^2 =& 3 \\ \\ x^2 + x^2 + x^2 =& 3 \\ \\ 3x^2 =& 3 \\ \\ x^2 =& 1 \\ \\ x =& \pm 1 \end{aligned} \end{equation}$


Therefore,

The other point of intersection is at $x = 1$

b.) Sketch the ellipse and the normal line.