1. Focal length

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

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?

2. Re: Focal length

Originally Posted by Melody2
hi,
I have to find the focal length of $\displaystyle x^2-6x+8y-7=0$

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?
There is a fast way. If you can get your equation into the form $\displaystyle y=a x^2 + b x + c$ then the focal length is $\displaystyle f=\frac{1}{4a}$

in this case you have to rearrange things a bit to obtain

$\displaystyle y=-\frac{1}{8}\left(x^2-6x-7\right)$ and thus read off $\displaystyle a$ as $\displaystyle -\frac{1}{8}$ so $\displaystyle f=-2$

3. Re: Focal length

nvm

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what is focal length in algebra 2

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