Answered

Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at
y = ± [tex]\frac{5}{6} x[/tex]


Sagot :

Answer:

[tex]\frac{y^2}{100} -\frac{x^2}{144} =1[/tex]

Step-by-step explanation:

By looking at the characteristics given, it can be determined that the equation will be vertical.

Being that the center is the midpoint of the vertices, the center is (0,0).

We have determined that the equation is vertical, so the equation used to find the asymptotes is [tex]y= k+\frac{a}{b} (x-h)\\ \\y=k-\frac{a}{b} (x-h)[/tex] . ± [tex]\frac{5}{6}[/tex] x is the asymptote.

We know that a is equal to 10 because of the vertices, so a squared is equal to 100.

[tex]\frac{a}{b} =\frac{5}{6}[/tex]

[tex]\frac{5}{6} =\frac{10}{b}[/tex]

When you cross multiply, and solve for b, you should get 12. Because b is 12, b squared is 144.

Now just fill the vertical transverse axis equation.

[tex]\frac{y^2}{100} -\frac{x^2}{144} =1[/tex]