Answer
The points on the ellipse are $~~(-\sqrt{\frac{2}{3}}, \frac{\sqrt{6}}{6})~~$ and $~~(\sqrt{\frac{2}{3}}, -\frac{\sqrt{6}}{6})~~$
Work Step by Step
$x^2+2y^2 = 1$
We can find $\frac{dy}{dx}$:
$x^2+2y^2 = 1$
$2x+4y\frac{dy}{dx} = 0$
$4y\frac{dy}{dx} = -2x$
$\frac{dy}{dx} = -\frac{2x}{4y}$
$\frac{dy}{dx} = -\frac{x}{2y}$
The tangent line has a slope of $1$ at points on the ellipse where $\frac{dy}{dx} = 1$
We can find an expression for $y$ when $\frac{dy}{dx} = 1$:
$\frac{dy}{dx} = -\frac{x}{2y} = 1$
$2y = -x$
$y = -\frac{x}{2}$
We can find the x-coordinates of the points on the ellipse that satisfy this condition:
$x^2+2y^2 = 1$
$x^2+2(-\frac{x}{2})^2 = 1$
$x^2+\frac{x^2}{2} = 1$
$\frac{3x^2}{2} = 1$
$x^2 = \frac{2}{3}$
$x = \pm \sqrt{\frac{2}{3}}$
When $x = -\sqrt{\frac{2}{3}}$:
$y = -\frac{x}{2}$
$y = -\frac{-\sqrt{\frac{2}{3}}}{2}$
$y = \sqrt{\frac{1}{4}}\cdot \sqrt{\frac{2}{3}}$
$y = \sqrt{\frac{2}{12}}$
$y = \sqrt{\frac{1}{6}}$
$y = \frac{\sqrt{6}}{6}$
When $x = \sqrt{\frac{2}{3}}$:
$y = -\frac{x}{2}$
$y = -\frac{\sqrt{\frac{2}{3}}}{2}$
$y = -\sqrt{\frac{1}{4}}\cdot \sqrt{\frac{2}{3}}$
$y = -\sqrt{\frac{2}{12}}$
$y = -\sqrt{\frac{1}{6}}$
$y = -\frac{\sqrt{6}}{6}$
The points on the ellipse are $~~(-\sqrt{\frac{2}{3}}, \frac{\sqrt{6}}{6})~~$ and $~~(\sqrt{\frac{2}{3}}, -\frac{\sqrt{6}}{6})~~$
