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So here's the problem:

Find the slope of the tangent line of : $2(x^2 +y^2)^2 = 25(x^2 - y^2)$ at the point (3,1)

Cool: So here's what I did:

Simplification step: $2(x^2 +y^2)^2 = 25(x^2 - y^2) \Rightarrow 2(x^2+y^2)^2 = 25x^2 - 25y^2$

Differentiate both sides: $4(x^2 +y^2)(2x(x') + 2y(y')) = 50x(x') - 50y(y')$


$(4x^2 + 4y^2)(2x + 2y(y'))= 50x - 50y(y')$

$(4x^2 + 4y^2)(2x + 2y(y'))= 50x - 50y(y')$

$8x^3 + 8x^2y(y') + 8xy^2 + 8y^3(y')= 50x - 50y(y')$

$8x^2y(y') + 8y^3(y') + 50y(y') = 50x - 8x^3 + 8xy^2$

$(y') (8x^2y + 8y^3 + 50y) = 50x - 8x^3 + 8xy^2$

$y' = \frac{50x - 8x^3 + 8xy^2}{8x^2y + 8y^3 + 50y}$

Cool, got y'...

Now, should I take the original equation and solve for y and substitute? I'm not even going to try putting down my steps on here.

The prompt is to find m=?

Someone please help?

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Just plug in the given $x,y$. It is better not to simplify. Differentiate immediately. Also, I wouldn't bother solving for general $y'$. As soon as you have differentiated, substitute. Then solve. – André Nicolas Oct 3 '13 at 2:43
as André suggests, you have $x=3$, you have $y=1$. So in your formula for $y'$ you can "Plug and chug" – Rustyn Oct 3 '13 at 2:44
up vote 2 down vote accepted

We start from $$2(x^2+y^2)^2=25(x^2-y^2).$$ Don't hesitate, Differentiate. We get $$4(x^2+y^2)(2x+2yy')=50x-50yy'.$$ Substitute $x=3$, $y=1$. We get $$4(10)(6+2m)=150-50m,$$ and therefore $130m=-90$.

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the equation is actually $2(x^2 + y^2)^2 = 25(x^2 - y^2)$, but thank you! – alvonellos Oct 3 '13 at 3:00
I bombed on my simplification, didn't I? – alvonellos Oct 3 '13 at 3:01
Yes, I noticed that, minor modification. – André Nicolas Oct 3 '13 at 3:01
Bombing on simplification happens a lot, at least to me. That's a very good reason not to do it if you don't have to. – André Nicolas Oct 3 '13 at 3:02
Thank you so much – alvonellos Oct 3 '13 at 3:03

$$y' = \frac{50x - 8x^3 + 8xy^2}{8x^2y + 8y^3 + 50y}$$ Substitute $x=3,y=1$ $$y' = \frac{50(3) - 8(3)^3 + 8(3)(1)^2}{8(3)^2(1) + 8(1)^3 + 50(1)}$$ $$y' = \frac{150 - 216 + 24}{72 + 8 + 50}$$ $$y' = \frac{-42}{130}$$ $$y' = -\frac{21}{65}$$

The slope of the line which is tangent to $2(x^2 +y^2)^2 = 25(x^2 - y^2)$ passes through point $(3,1)$ is $-\frac{21}{65}$

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It says that's wrong. – alvonellos Oct 3 '13 at 2:54

The answers above were wrong since the equation is wrong. It should be the following - y1= 8(x^2+y^2)-50)x/-(8(x^2+y^2)+50)y Then you get the slope as -9/13.

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Can you provide a reason for why you say that the answers are wrong? – user61527 Oct 31 '13 at 22:07

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