Question

Write the equation of the hyperbola with vertices (2,3) and (2,9) and co-vertex (7,6)?

Answer 1

Please help! I would appreciate any and all help on this question. I know you would use the formula x^2 / a^2 - y^2 / b^2, but how would I input the variables?

Answer 2

Vertices of the hyperbola (2, 3) and (2, 9) lie on the same straight, vertical line x = 2. Thus the hyperbola is a conjugate hyperbola. This means that instead of the formula x^2/a^2 - y^2/b^2, you will use the formula -x^2/a^2 + y^2/b^2.

Since the vertices of the hyperbola are (2, 3) and (2, 9), hence its center is the midpoint of these, that is, ((2 + 2)/2, (3 + 9)/2) = (2, 11/2)

Hence the equation of the hyperbola is
\[-\frac{ (x - 2)^2 }{ a^2 } + \frac{(y - 11/2)^2}{b^2} = 1\]

Can you proceed from here?

Answer 3

@navk - Thank you so much for the help!
but how would I get a and b? I thought a would equal 2?

Answer 4

Length of transverse axis for a conjugate hyperbola is 2b. Since the vertices are (2, 3) and (2, 9) hence distance between them is 9 - 3 = 6.

Thus 2b = 6 and b = 3

Answer 5

*Error: The center of the ellipse, as stated in the first post, is not (2, 11/2) but (2, 6), so the equation of the ellipse is:

\[-\frac{ (x-2)^2 }{ a^2 } + \frac{(y-6)^2}{b^2} = 1 \]

We know that b = 3.

Now co-vertex is (7, 6). It's distance from the center (2, 6) is 7 - 2 = 5. This means that a = 5. (since 2a is the length of the conjugate axis)

Hence putting a and b into the equation

\[-\frac{ (x-2)^2 }{ 25 } + \frac{(y-6)^2}{9} = 1 \]

Answer 6

@navk - Thanks a lot!

Answer 7

you're welcome