Question

hyperbolas

Answer 1

find an equation in standard form for the hyperbola with vertices at (0.+-4) and asymptotes at y=+-2/3x

Answer 2

chart was not helpful ?

Answer 3

I think the first step would be to figure out whether the required hyperbola is horizontal or vertical

Answer 4

so its y^2/4^2 -x^2/... idk what that ones over .-.

Answer 5

try to sketch it

Answer 6

Excellent ! look at the last row in the chart

Answer 7

asymptotes are \(\large y = \pm\dfrac{a}{b} x =\pm\dfrac{2}{3}x \)

Answer 8

\(\large \implies \dfrac{a}{b} = ?\)

Answer 9

oh so b is 3 so its y^2/4-x^2/9

Answer 10

thats clever :) but no !

Answer 11

\(\large \dfrac{a}{b} = \dfrac{2}{3}\)

Answer 12

you already knw the value of \(\large a\), so plug it in and solve \(\large b\)

Answer 13

but :c b it 3..... it says so right there a/b=2/3.....
so its y^2/9-x^2/2

Answer 14

\(\large a = 4\) since the vertices are (0, +-4)

\(\large \dfrac{4}{b} = \dfrac{2}{3} \implies b = 6\)

Answer 15

the equation would be \(\large \dfrac{y^2}{4^2} - \dfrac{x^2}{6^2} = 1\)

Answer 16

okay so itd be y^2/16-x^2/36=1

Answer 17

yes.. and you may verify it quick by finding the vertices and asymptotes (use the chart)