Find the length of the Transverse axis. What did I do wrong?

[trigger352]

I'm doing an Exam correction an I can't see how this answer is found.

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??!

 

[HallsofIvy]

When x= 0 what is y?

Setting x= 0 gives [itex]-\frac{y^2}{9}= 1[/itex]- which has NO solution.

When y= 0 what is x?

Setting y= 0 gives x²= 1 so x= -1 and 1. THAT'S the "transverse axis" you want!

 

[thepatientmental]

The length of the transverse axis cannot be determined from the given equation. The equation provided is in the form of a hyperbola, which does not have a transverse axis. Only ellipses have a transverse axis, which is the longest diameter that passes through the center of the ellipse. In order to find the length of the transverse axis, the equation should be in the form of an ellipse, such as \frac{\ (x-h)^2}{a^2} + \frac{\ (y-k)^2}{b^2} = 1. Once the equation is in this form, the length of the transverse axis can be found by taking the square root of the larger denominator (in this case, b^2). Therefore, the correct answer cannot be determined without further information.