Question

Find the equation of tangent to curve `y^2(y^2-4)=x^2(x^2-5)` at point `(0,-2)`?

Answer 1

Equation of tangent at `(0,-2)` is `y+2=0`

Explanation:

Let us find the derivative `(dy)/(dx)` for `y^2(y^2-4)=x^2(x^2-5)`, using implicit differentiation. The differential is

`y^2xx2yxx(dy)/(dx)+2yxx(y^2-4)xx(dy)/(dx)=x^2xx2x+2x xx(x^2-5)`

or `2y^3(dy)/(dx)+(2y^3-8y)(dy)/(dx)=2x^3+(2x^3-10x)`

or `(dy)/(dx)=(4x^3-10x)/(4y^3-8y)`

and at `x=0` and `y=-2`, `(dy)/(dx)=0`

As slope of tangent is `0` and it passes through `(0,-2)`

the equation of tangent is `(y+2)=0(x-0)` or `y+2=0`

graph{(y+2)(y^2(y^2-4)-x^2(x^2-5))=0 [-10, 10, -5, 5]}