# Thread: Quick question

1. ## Quick question

$y^2-2x^2 = 2y+1$
Need to find all points where the tangent line is horizontal.
To do this you just find the derivative and set it to 0 and solve right? The only thing that is tripping me up is the equation. What form do I need to put it into?

2. Originally Posted by zsig013
$y^2-2x^2 = 2y+1$
Need to find all points where the tangent line is horizontal.
To do this you just find the derivative and set it to 0 and solve right? The only thing that is tripping me up is the equation. What form do I need to put it into?
Use implicit differentiation to get dy/dx as a function of y and x: $\frac{dy}{dx} = \frac{2x}{y-1}$.

Solve $\frac{dy}{dx} = 0$: x = 0.

Substitute x = 0 into $y^2-2x^2 = 2y+1$ and solve for the y-coordinate. Check that the solution(s) do NOT satisfy y - 1 = 0 (why?)

3. ## Ok

Originally Posted by zsig013
$y^2-2x^2 = 2y+1$
Need to find all points where the tangent line is horizontal.
To do this you just find the derivative and set it to 0 and solve right? The only thing that is tripping me up is the equation. What form do I need to put it into?
Another way to do this is to solve for $y$ in you original equation to get $y=\pm\sqrt{2(x^2+1)}+1$...so you get $y'=\frac{\pm{\sqrt{2}x}}{\sqrt{x^2+1}}$...so $y'=0,iff x=0$...now using this you can come to teh same conclusing Mr. Fantastic did...this is a little harder but a neat way of doing it