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Question

Find an equation of the tangent line to the given curve $2\left(x^2+y^2\right)^2=25\left(x^2-y^2\right)$ at the point $(3,1)$.

Solution

VerifiedAnswered 2 years ago

Answered 2 years ago

Step 1

1 of 7Differentiating both sides of the equation with respect to $x$, we obtain

$\begin{aligned} \dfrac{d}{dx}[2(x^2+y^2)^2]&=\dfrac{d}{dx}[25(x^2-y^2)]\\[0.7em] 4(x^2+y^2)\cdot\dfrac{d}{dx}(x^2+y^2)&=25\dfrac{d}{dx}(x^2-y^2)\\[0.7em] 4(x^2+y^2)\Big(2x+2y\dfrac{dy}{dx} \Big)&=25\Big(2x-2y\dfrac{dy}{dx} \Big) \end{aligned}$

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