## Calculus 8th Edition

$y=-2$
$y^2(y^2-4)=x^2(x^2-5)\\ (y^2)(2yy')+(y^2-4)(2yy')=(x^2)(2x)+(x^2-5)(2x)\\ 2y^3y'+2y^3y'-8yy'=2x^3+2x^3-10x\\ 4y^3y'-8yy'=4x^3-10x\\ 2y'(2y^3-4y)=2(2x^3-5x)\\ y'=\frac{2x^3-5x}{2y^3-4y}$ Plug in $(0,-2)$ into $y′$ to find the gradient $y'=\frac{2(0)^3-5(0)}{2(-2)^3-4(-2)}\\ y'=0$ Equation of the tangent line at $(0,-2)$ $y-(-2)=0(x-0)\\ y=-2$