I'm doing an Exam correction an I can't see how this answer is found.(adsbygoogle = window.adsbygoogle || []).push({});

Question: [tex]x^2 - \frac{\ y^2}{9} = 1[/tex]

Find the length of the transverse axis.

I took [tex]x^2 - \frac{\ y^2}{9} = 1[/tex]

And plugged it into

[tex]\frac{\ (x-h)^2}{b^2} - \frac{\ (y-k)^2}{a^2} = 1[/tex]

to get

[tex]\frac{\ (x-0)^2}{1^2} - \frac{\ (y-0)^2}{3^2} = 1[/tex]

The 0's were because in the original equation, I didn't see an [tex]h[/tex] or [tex]k[/tex] value that was affecting either [tex]x[/tex] or [tex]y[/tex]

The [tex]1^2[/tex] is because [tex]x[/tex] = [tex]\frac{x}{1}[/tex] like [tex]a[/tex] = [tex]\frac{a}{1}[/tex]

[tex]a = 3[/tex], therefore the length of the transverse is 6 units...

....Incorrect Answer??!

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# Homework Help: Find the length of the Transverse axis. What did I do wrong?

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