Tangent line with a slope of 1 to $x^2+2y^2=1$ I had a problem on the test I just took that I have never seen before.
$x^2 + 2y^2 = 1$ and  was suppose to find the tangent lines on that curve that have a slope of one.
I just couldn't figure out how to do it. I am not even sure if I did anything but I got the derivative as
$2x + 4y y\prime$ and then from there I did some algera but I don't think any of that was correct.

Basically, you want to find $(x,y)$ such that
$$\frac{dy}{dx}=1$$
and
$$x^2+2y^2=1$$
You have correctly put
$$2x +4 y \frac{dy}{dx} =0$$
Now you solve for $y'$, and get
$$ - \frac{x}{{2y}} = \frac{{dy}}{{dx}}$$
So, since you're looking for $1 = \dfrac{{dy}}{{dx}}$, you need:
$$ - \frac{x}{{2y}} = 1$$
or
$$-x = 2y$$
Squaring the equation gives:
$$x^2 = 4y^2$$
$$\frac{x^2}{2} = 2y^2$$
Substituting in our original equation you have:
$$2y^2 + {x^2} = 1$$
$$\frac{{{x^2}}}{2} + {x^2} = 1$$
this yields $x=\pm \sqrt{\dfrac{2}{3}}$
These values actually produce $|y'|=1$ so you need to choose the $y$ coordinate appropriately. See by yourself:

