find the equation of the tangent line to the curve (a lemniscate) ( 2(x^{2}+y^{2})^{2}=25(x^{2}-y^{2}) ) at…

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})