Originally Posted by

**Melody2** hi,

I have to find the focal length of $\displaystyle x^2-6x+8y-7=0$

I had already found the turning point using calculus. (3,2)

I've never done it from this starting point but I rearranged it to get this

$\displaystyle \begin{align*}x^2-6x+9&=-8x-16\\(x-3)^2&=-8(y-2)\\(x-3)^2&=4\times-2(y-2)\end{align*}$

which reaffirms that the vertex is (3,2) and gives the focal length of a=-2

Is this the BEST way of doing it, especially considering that I had already used calculus to find the vertex?