Question

use implicit differentiation to find an equation of the tangent line to the curve at the given point.

$2(x^{2}+y^{2})^{2}=25(x^{2}-y^{2})$, $(3,1)$

Explanation:

Step1: Differentiate both sides with respect to \(x\)

Differentiate \(2(x^{2}+y^{2})^{2}\) using the chain - rule. Let \(u = x^{2}+y^{2}\), then \(\frac{d}{dx}(2u^{2})=4u\frac{du}{dx}\), and \(\frac{du}{dx}=2x + 2y\frac{dy}{dx}\). So \(\frac{d}{dx}(2(x^{2}+y^{2})^{2})=4(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})\).

Differentiate \(25(x^{2}-y^{2})\) with respect to \(x\): \(\frac{d}{dx}(25(x^{2}-y^{2}))=25(2x-2y\frac{dy}{dx})\).

So we have \(4(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})=25(2x - 2y\frac{dy}{dx})\).

Step2: Substitute the point \((3,1)\)

First, find \(x = 3\) and \(y = 1\). Then \(x^{2}+y^{2}=3^{2}+1^{2}=9 + 1=10\).

Substitute into \(4(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})=25(2x - 2y\frac{dy}{dx})\):

\(4\times10\times(2\times3+2\times1\times\frac{dy}{dx})=25\times(2\times3 - 2\times1\times\frac{dy}{dx})\)

\(40\times(6 + 2\frac{dy}{dx})=25\times(6-2\frac{dy}{dx})\)

Expand: \(240+80\frac{dy}{dx}=150 - 50\frac{dy}{dx}\)

Move terms with \(\frac{dy}{dx}\) to one side: \(80\frac{dy}{dx}+50\frac{dy}{dx}=150 - 240\)

\(130\frac{dy}{dx}=-90\), so \(\frac{dy}{dx}=-\frac{9}{13}\).

Step3: Use the point - slope form \(y - y_{1}=m(x - x_{1})\)

Here \(x_{1}=3,y_{1}=1,m =-\frac{9}{13}\)

\(y - 1=-\frac{9}{13}(x - 3)\)

\(y-1=-\frac{9}{13}x+\frac{27}{13}\)

\(y=-\frac{9}{13}x+\frac{27 + 13}{13}\)

\(y=-\frac{9}{13}x+\frac{40}{13}\)

Answer:

\(y =-\frac{9}{13}x+\frac{40}{13}\)