Find an equation for the hyperbola described?

Question

Vertices at (0, ±10), asymptotes at y=±(3/5)x

Answer with clear explanation preferred!

Answer 1

The center of the hyperbola is the midpoint between the two vertices.

((0+0)/2,(10-10)/2) = (0,0)

The vertices run vertically so the hyperbolas open vertically. So the equation is of the form:

y²/a² - x²/b² = 1

The distance between the two vertices is 2a.
2a = 10 - (-10) = 10 + 10 = 20
a = 10
a² = 100

The equation of the asymptotes is y = ±(3/5)x

Since the hyperbolas open vertically the slope
m = ±a/b = ±10/b = ±3/5
±10*5 = ±3b
±50/3 = b
b² = 2500/9

The equation of the hyperbola is:

y²/a² - x²/b² = 1
y²/100 - x²/(2500/9) = 1
y²/100 - 9x²/2500 = 1

Answer 2

Let y = ± (3/5) x.... then

y² = (9/25) x² + C

At x = 0, y = ± 10, so that C = 100, so that the final equation is:

25 y² - 9 x² + 2500 = 0

Note: To find equation of asymptotes, ignore all powers of y and x less the highest ones. In this case, it becomes 25 y² = 9 x² so that it ends up being y = ± (3/5) x. This is a rule that works for any equation that has asymptotes, not just hyperbolas, so it's worth it for you to understand this, rather try to remember some stupid rule "for hyperbolas" because you're going to forget those rules. Even I don't remember the rule "for hyperbolas", it's a joke.

Answer 3

center is origin(mid poit of vertices)
vertives lie on y axis hence equation is of form
y^2/b^2 - x^2/a^2=1
b=10
equation of assymptotes : y=+-b/a
b/a=3/5
a=50/3