Question

Find an equation of the tangent line to the graph at the given point. ?

Answer 1

`y = sqrt{3}/6 x-{8sqrt{3}}/3`

Explanation:

Differentiating both sides of the implicit equation
`x^2y^2 -9x^2-4y^2 =0`
with respect to `x` leads to
`2x^2 y {dy}/{dx} + 2xy^2 -18x - 8y {dy}/{dx} = 0`
or
`2y(x^2-4) {dy}/{dx} +2xy^2-18x = 0`
Substituting `x=4` and `y=-2sqrt{3}` leads to
`dy/dx = {sqrt{3}}/6`
Since this is the slope of the tangent line passing through `(4,-2sqrt{3})`, the equation of the tangent can be found :
`y-(-2sqrt{3}) = {sqrt{3}}/6 (x-4)`