Implicit Derivative at a Point

Question

If -3y^2 - x^3 = -xy^3 - 1 then find the equations of all tangent lines to the curve when x = 1

Answer 1

If we have -3y²-x³=-xy³-1, at x=1 we have

-3y²-1=-y³-1

Setting this up for solving we get

y³-3y²=0 which forces y= or y=3. We have the point (1,0) and (1,3).

By implicit differentiation we get:

-6y(y')-3x²=-y³-3xy²(y') at (1,0) we get -3=0 which is a false statement. Graphically we see a vertical tangent line at (1,0) so this supports we have x=1.

At (1,3) we get

-18y'-3=-27-27y' which forces y'=-8/3

So my second tangent line in point slope form is: y-3=-8/3(x-1)