Question

Find an equation of the tangent line to the given curve 2(x2+y2)2=25(x2y2)2\left(x^2+y^2\right)^2=25\left(x^2-y^2\right) at the point (3,1)(3,1).

Solution

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Answered 2 years ago
Answered 2 years ago
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Differentiating both sides of the equation with respect to xx, we obtain

ddx[2(x2+y2)2]=ddx[25(x2y2)]4(x2+y2)ddx(x2+y2)=25ddx(x2y2)4(x2+y2)(2x+2ydydx)=25(2x2ydydx)\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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