Question

Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at
y = ±5/4 x.

Answer 1

center is at the origin, half way between the vertices, so you know it looks like
\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\]

Answer 2

\(a=10\implies a^2=100\) so now we have
\[\frac{x^2}{100}-\frac{y^2}{b^2}=1\]

Answer 3

for \(b\) note that \(\frac{a}{b}=\frac{10}{b}=\frac{5}{4}\) and solve for \(b\)

Answer 4

b=12, y^2/100-x^2/144? @satellite73

Answer 5

\[\large \frac{10}{b}=\frac{5}{4}\] cross multiply..

Answer 6

okay, wrong number..opps. b5=14.. 11?did i do that right?

Answer 7

@Jhannybean

Answer 8

what is 4 times 10?

Answer 9

\[\large 10 + 4 \ne 10 \cdot 4\]

Answer 10

40..

Answer 11

and what is 40 divided by 5?

Answer 12

8

Answer 13

we're getting there right?

Answer 14

yeah, It's been a really long day..haha

Answer 15

so a is 5 and b is 8?

Answer 16

i can tell
you should solve
\[\frac{10}{b}=\frac{5}{4}\] in your head by noting that \(10=2\times 5\) and so \(b=2\times 4\)

Answer 17

i.e. \(b=8\) and so the equation is
\[\frac{x^2}{100}-\frac{y^2}{64}=1\] lets check it

Answer 18

oops i had it backwards !!

Answer 19

\[\frac{y^2}{100}-\frac{x^2}{64}=1\]

Answer 20

that's better
http://www.wolframalpha.com/input/?i=hyperbola+y^2%2F100-x^2%2F64%3D1

Answer 21

okay I understand now thanks guys!