Question

Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at y = ±(3/7)x HELP ME :(

Answer 1

@AkashdeepDeb do you know how to do this one by any chance too? :l

Answer 2

I don't know what asymptotes are.
Sry....
Tag some people!! :D

Answer 3

alright np :D thank you!!!

Answer 4

@phi @E.ali

Answer 5

See http://www.mathsisfun.com/geometry/hyperbola.html

Answer 6

If you scroll down the "Equation" it shows a hyperbola facing sideways. ) (
Your hyperbola is a "smile/frown", so you have to swap x and y in their example.

That is, your hyperbola's equation is
\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

Answer 7

according to the site,
One vertex is at (a, 0), and the other is at (-a, 0)
or in this case, where x and y are swapped

One vertex is at (0,a), and the other is at (0,-a)

so what is "a" for your hyperbola ?

Answer 8

ahh okay I see, so a is +-6 right?

Answer 9

yes, so if we replace a^2 with 36, the equation is
\[ \frac{y^2}{36} - \frac{x^2}{b^2} = 1 \]

we need to find b to finish this.

Answer 10

cool! wait how do we find b??

Answer 11

Again from the site we have (for a *sideways*) hyperbola
The asymptotes are the straight lines:
y = (b/a)x
y = -(b/a)x

we have to swap x and y for this problem (an up/down hyperbola)
x = (b/a) y
x = (-b/a)y

they tell you y = ±(3/7)x
if we look at the +, we have y = (3/7)x
or, solving for x, x= (7/3)y
match that equation with x= (b/a)y
with a=6, we have x= (b/6)y

so we have x= (7/3)y and x = (b/6)y

we can set (7/3)y =(b/6)y
can you find b ?

Answer 12

14 right? and then I'd square that under x^2 to get 196?

Answer 13

to solve
\[ \frac{7}{3} y = \frac{b}{6} y \]
first divide both sides by y. You get
\[ \frac{7}{3} \cdot \frac{y}{y} = \frac{b}{6} \cdot\frac{y}{y} \\ \frac{7}{3} = \frac{b}{6}\]
multiply both sides by 6

Answer 14

yes, b=14
and the equation is
\[ \frac{y^2}{36} - \frac{x^2}{196} = 1\]

Answer 15

AHH THANK YOU SO MUCH :) it's a lot clearer now.

Answer 16

See
http://www.wolframalpha.com/input/?i=%28y%2F6%29%5E2+-+%28x%2F14%29%5E2+%3D+1

where it says, Geometric Figure: hyperbola, click on Properties on the right side to get a list of properties... check that the asymptotes and vertex match with your problem