Question

find the equation of the tangent line to the curve (a lemniscate) ( 2(x^{2}+y^{2})^{2}=25(x^{2}-y^{2}) ) at the point ( (3,1) ). the equation of this tangent line can be written in the form ( y = mx + b ) where ( m ) is: and where ( b ) is:

Explanation:

Step1: Differentiate both sides of the equation

Differentiate \(2(x^{2}+y^{2})^{2}=25(x^{2}-y^{2})\) with respect to \(x\) using the chain - rule and product - rule.

Let \(u = x^{2}+y^{2}\), then \(\frac{d}{dx}(2u^{2})=4u\frac{du}{dx}\), and \(\frac{du}{dx}=2x + 2y\frac{dy}{dx}\).

For the right - hand side, \(\frac{d}{dx}(25(x^{2}-y^{2}))=25(2x-2y\frac{dy}{dx})\)

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

Step2: Substitute \(x = 3\) and \(y = 1\)

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

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

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

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

Step3: Solve for \(\frac{dy}{dx}\) (which is \(m\))

Add \(50\frac{dy}{dx}\) to both sides: \(240+80\frac{dy}{dx}+50\frac{dy}{dx}=150\)

\(240 + 130\frac{dy}{dx}=150\)

Subtract \(240\) from both sides: \(130\frac{dy}{dx}=150 - 240=-90\)

\(m=\frac{dy}{dx}=-\frac{9}{13}\)

Step4: Find \(b\)

We know the equation of the line is \(y=mx + b\), substituting \(x = 3\), \(y = 1\) and \(m=-\frac{9}{13}\)

\(1=-\frac{9}{13}\times3 + b\)

\(1=-\frac{27}{13}+b\)

\(b=1+\frac{27}{13}=\frac{13 + 27}{13}=\frac{40}{13}\)

Answer:

\(m =-\frac{9}{13}\), \(b=\frac{40}{13}\)