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
