Question

Help please! How to find length of transverse axis and conjugate axis?

Answer 1

(x-4)^2/36 - (y-2)^2/9=1

Answer 2

@ranga

Answer 3

@ranga I have to leave for a bit, but if you could post the answer and EXPLAIN how you got the answer I would appreciate it. I know everything else like asymptotes, vertices, center and such, but just not these two. Thanks for your help!

Answer 4

length traverse axis --> another name for "major axis"
conjugate axis --> another name for "minor axis"

Answer 5

Hyperbola in conics form: (x-h)^2 / a^2 - (y-k)^2 / b^2 = 1

Length of transverse axis = 2a

Length of conjugate axis = 2b

You can compare (x-4)^2 / 36 - (y-2)^2 / 9 = 1 to
(x-h)^2 / a^2 - (y-k)^2 / b^2 = 1

and quickly identify a and b.

Answer 6

\(\bf \cfrac{(x-4)^2}{36}-\cfrac{(y-2)^2}{9}=1\implies \cfrac{(x-4)^2}{6^2}-\cfrac{(y-2)^2}{3^2}=1\\ \quad \\\implies \cfrac{(x-h)^2}{a^2}-\cfrac{(y-k)^2}{b^2}=1\)

Answer 7

so transverse=12 and conjugate=6?

Answer 8

yeap