Question

Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at
y = ±five divided by six times x.

Answer 1

so the vertices are at (0, 10) and (0, -10)
notice the "x" coordinate didn't change, the "y" did
that means it has a vertical traverse axis, or is vertically opening
it moved up 10 units and down 10 units, meaning the center is a the origin
a = 10

the asymptotes form for a vertical hyperbola are at
$$\bf y= k\pm \cfrac{a}{b}(x-h)\\
\cfrac{a}{b} \text{ being the slope for the asympote equation}\\
\text{so your asymptotes are at }y =\pm\cfrac{5}{6}x\\
\text{notice the slope of the equation, center is (0, 0)}\\
k\pm \cfrac{a}{b}(x-h) \implies 0\pm \cfrac{a}{b}(x-0)\implies \pm \cfrac{a}{b}x \implies \pm\cfrac{5}{6}x
$$

Answer 2

we know that a = 10, so we can say by equating both slopes that
$$ \bf \pm \cfrac{a}{b}= \pm\cfrac{5}{6}\\
\cfrac{a}{b}= \cfrac{5}{6}\\
\large or\\
-\cfrac{a}{b}= -\cfrac{5}{6}
$$
solve either of those 2 slope equations, to get the "b" component

Answer 3

woops, forgot one thing :/

Answer 4

$$\bf \pm \cfrac{10}{b}= \pm\cfrac{5}{6}\\
\cfrac{10}{b}= \cfrac{5}{6}\\
\large or\\
-\cfrac{10}{b}= -\cfrac{5}{6}
$$

Answer 5

Thank you!!

Answer 6

yw

Answer 7

Is the final answer: y^2/64 - x^2/100?
Since b = 8