Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 54

Determine the points on the ellipse $x^2+2y^2=1$ where the tangent line has slope 1.
Solving fro $x$, where slope$(m) = 1 = y'$
$\displaystyle \frac{d}{dx} (x^2) + 2 \frac{d}{dx}(y^2) = \frac{d}{dx}(1)$

$\begin{equation} \begin{aligned} 2x + 4y \frac{dy}{dx} &= 0\\ \\ 5yy' &= -2x\\ \\ \frac{\cancel{4y}y'}{\cancel{4y}} &= \frac{-2x}{4y}\\ \\ y' &= \frac{-x}{2y}\\ \\ 1 &= \frac{-x}{2y}\\ \\ 2y &= -x\\ \\ \frac{\cancel{2}y}{\cancel{2}} &= \frac{-x}{2}\\ \\ y &= \frac{-x}{2} \end{aligned} \end{equation}$

Using the given equation


$\begin{equation} \begin{aligned} x^2 + 2y^2 &= 1\\ \\ x^2 + 2 \left( \frac{-x }{2} \right)^2 &= 1\\ \\ x^2 + 2 \left( \frac{-x }{4} \right) &= 1\\ \\ x^2 + \frac{x^2}{2} &= 1\\ \\ \frac{2x^2 + x^2}{2} &= 1 \\ \\ \frac{3x^2}{2} &= 1\\ \\ \frac{\cancel{2}}{\cancel{3}} \left[ \frac{\cancel{3}}{\cancel{2}}\right. x^2 &= \left.\phantom{\frac{\cancel{3}}{\cancel{2}}} 1 \right] \frac{2}{3}\\ \\ x^2 &= \frac{2}{3}\\ \\ \sqrt{x^2} &= \pm \sqrt{\frac{2}{3}}\\ \\ x^2 &= \pm \sqrt{\frac{2}{3}} \end{aligned} \end{equation}$


Solving for $y$, using the given equation
@ $\displaystyle x = \pm \frac{2}{3}$

$\begin{equation} \begin{aligned} \left( \sqrt{\frac{2}{3}} \right)^2 + 2y^2 &= 1\\ \\ \frac{2}{3} + 2y^2 &= 1\\ \\ 2y^2 &= 1 - \frac{2}{3}\\ \\ 2y^2 &= \frac{3-2}{3}\\ \\ 2y^2 &= \frac{1}{3}\\ \\ \frac{\cancel{2}y^2}{\cancel{2}} &= \frac{\frac{1}{3}}{2}\\ \\ y^2 &= \frac{1}{6}\\ \\ \sqrt{y^2} &= \pm \sqrt{\frac{1}{6}}\\ \\ y &= \pm \sqrt{\frac{1}{6}} \end{aligned} \end{equation}$


The points on the ellipse $x^2 + y^2 = 1$ where the slope is 1 are $\displaystyle \left( \pm \sqrt{\frac{2}{3}}, \pm \sqrt{\frac{1}{6}}\right)$