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

Answer:

Step-by-step explanation:

Given that:

[tex]y^2 (y^2-4) = x^2(x^2 -5)[/tex]

at point (0, -2)

[tex]\implies y^4 -4y^2 = x^4 -5x^2[/tex]

Taking the differential from the equation above with respect to x;

[tex]4y^3 \dfrac{dy}{dx}-8y \dfrac{dy}{dx}= 4x^3 -10x[/tex]

Collect like terms

[tex](4y^3 -8y)\dfrac{dy}{dx}= 4x^3 -10x[/tex]

[tex]\dfrac{dy}{dx}= \dfrac{4x^3 -10x}{4y^3-8y}[/tex]

Hence, the slope of the tangent line m can be said to be:

[tex]\dfrac{dy}{dx}= \dfrac{4x^3 -10x}{4y^3-8y}[/tex]

At point (0,-2)

[tex]\dfrac{dy}{dx}= \dfrac{4(0)^3 -10(0)}{4(-2)^3-8-(2)}[/tex]

[tex]\dfrac{dy}{dx}= \dfrac{0 -0}{4(-8)+16}[/tex]

[tex]\dfrac{dy}{dx}= 0[/tex]

m = 0

So, we now have the equation of the tangent line with slope m = 0 moving through the point (x, y) = (0, -2) to be:

(y - y₁ = m(x - x₁))

y + 2 = 0(x - 0)

y + 2 = 0

y = -2

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