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: $4(x^2 +y^2)(2x(x') + 2y(y')) = 50x(x') - 50y(y')$
Simplify:
$(4x^2 + 4y^2)(2x + 2y(y'))= 50x - 50y(y')$
$(4x^2 + 4y^2)(2x + 2y(y'))= 50x - 50y(y')$
$8x^3 + 8x^2y(y') + 8xy^2 + 8y^3(y')= 50x - 50y(y')$
$8x^2y(y') + 8y^3(y') + 50y(y') = 50x - 8x^3 + 8xy^2$
$(y') (8x^2y + 8y^3 + 50y) = 50x - 8x^3 + 8xy^2$
$y' = \frac{50x - 8x^3 + 8xy^2}{8x^2y + 8y^3 + 50y}$
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?
$\endgroup$ 22 Answers
$\begingroup$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 $130m=-90$.
$\endgroup$ 5 $\begingroup$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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