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Find an equation of the tangent line to the curve $2·(x^2+y^2)^2=25·(x^2−y^2)$ at the point $(3,1)$.

Any hints?

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  • $\begingroup$ You can use implicit derivation $\endgroup$ Commented Nov 8, 2014 at 18:09
  • $\begingroup$ $y^{'\:}=-\frac{x\left(4y^2+4x^2-25\right)}{y\left(4y^2+4x^2+25\right)}$ The what do I do? $\endgroup$
    – Csci319
    Commented Nov 8, 2014 at 18:17

2 Answers 2

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The more general way to solve this type of problem is with implicit differentiation:

We have $2(x^2+y^2)^2=25(x^2-y^2)$. Differentiating both sides w.r.t. $x$, we get: $$4(x^2+y^2)(2x+2yy')=50x-50yy'$$ $$\implies (8(x^2+y^2)y+50y)y'=50x-4(x^2+y^2)\cdot 2x$$ $$\implies y'=\frac {50x-8x^3-8xy^2}{8x^2y+8y^3+50y}$$ $$\implies y'=-\frac{x(4x^2+4y^2-25)}{y(4x^2+4y^2+25}$$

Because we know that derivatives are just the slopes of tangent lines to graphs, we know that the slope of this implicit function at point $(3,1)$ is $y'|_{(3,1)}=-\frac{(3)(4(9)+4(1)-25)}{(1)(4(9)+4(1)+25)}=-\frac{9}{13}$.

We then just plug in that slope and point into the point-slope equation of a line: $$y-y_0=m(x-x_0)$$ and we have the answer.

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Expanding: $$4x^2+4y^2 = 25x^2-25y^2,$$ $$29y^2=21x^2.$$ Note that we can take the positive root since the point of interest is in the first quadrant. $$y=\sqrt{\frac{21x^2}{29}}.$$ From here you are able to solve the problem.

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