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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.

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    Looks like you're on the right track. Take the derivative of both sides, and you get $2x + 4y\ y' = 0$. You're trying to find where $y' = 1$, so plug that in, and you get $2x + 4y = 0$. Now combine this equation with the original equation, $x^2 + 2 y^2 = 1$, and you have a system of two equations with two unknowns.2012-03-08

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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:

enter image description here

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    @Jordan I guess you need to review some of your algebra topics then. Your problem is solved as follows: 1. We find a formula for $y'$. 2. Since we need $y'=1$, which is a function of $y=f(x)$ or $x$ and $y$, then wether $y'=1$ or not will depend on the values of $x$ and $y$, in this case, this will happen only when $-x=2y$. But this tells you "nothing" unless you consider that you also need $x$ and $y$ to be on the ellipse $x^2+2y^2=1$. It is by considering both equations that you obtain a solution. Make and effor, think it out, think it out, again. Try to solve it by yourself with this ideas.2012-03-08