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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. –  Tanner Swett Mar 8 '12 at 1:22
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1 Answer

up vote 4 down vote accepted

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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So, you need $$-x = 2y$$ or $$x^2 = 4y^2$$ I do not understand that part. –  user138246 Mar 8 '12 at 3:17
    
@Jordan If we have that $2=2$, doesn't that also mean that $2^2=2^2$? –  Pedro Tamaroff Mar 8 '12 at 3:21
    
I don't follow, -x = 2y could mean many things and I don't see why it is squared. –  user138246 Mar 8 '12 at 3:23
    
@Jordan What we need is to solve a system of equations. Namely $x^2=2y^2=1 \wedge -x=2y$. Since we can manipulate an equation (for example, square it), without breaking the inequality, we do so. Having $x^2=2\cdot 2 y^2$ is useful because we can substitute this in the first equation, finding $x$, and then finding $y$. Do you follow? –  Pedro Tamaroff Mar 8 '12 at 3:27
    
I thought I remembered hearing that you could not square both sides of an equation because that could change the value. –  user138246 Mar 8 '12 at 4:08
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