Question

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. y 2 ( y 2 − 4 ) = x 2 ( x 2 − 5 ) , ( 0 , − 2 ) (devil's curve) y2(y2-4)=x2(x2-5), (0,-2) (devil's curve)

Answer 1

Answer:

y = -2

Step-by-step explanation:

To find the equation of the tangent we apply implicit differentiation, and then we take apart dy/dx

The equation is

$y^2(y^2-4)=x^2(x^2-5)$

implicit differentiation give us

$\frac{d}{dx}[y^2(y^2-4)=x^2(x^2-5)]\\\\2y\frac{dy}{dx}(y^2-4)+y^2(2y\frac{dy}{dx})=2x(x^2-5)+x^2(2x)\\\\4y^3\frac{dy}{dx}-8y\frac{dy}{dx}=2x^3-10x+2x^3\\\\\frac{dy}{dx}=\frac{4x^3-10x}{4y^3-8y}$

But we know that

$m=\frac{dy}{dx}\\y=mx+b$

Hence, for the point (0,-2) and by replacing for dy/dx

$m=\frac{dy}{dx}_{(0,-2)}=\frac{4(0)+10(0)}{4(-2)^3-8(-2)}=0$

Hence m=0, that is, the tangent line to the point is a horizontal line that cross the y axis for y=-2. The equation is:

y=(0)x+b = -2

HOPE  THIS HELPS!!