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 $\displaystyle \left( \pm \sqrt{\frac{2}{3}}, \pm \sqrt{\frac{1}{6}}\right)$