Question

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

Answer 1

The equation of this hyperbola in standard form:
y² / a² + x² / b² = 1.
y = +/- a/b x
a / b = 5 / 4
a = 10
10 / b = 5 / 4
b = (10 · 4) : 5
b = 8
Answer:
The equation of the hyperbola is:
y² / 100 - x² / 64 = 1

Answer 2

Answer:

Step-by-step explanation:

Given that vertex of the hyperbola is

(0,10) and(0,-10)

Hence the hyperbola will have equation of the form

$\frac{y^2}{a^2} -\frac{x^2}{b^2} =1$

Since vertex has y coordinate as 10, we have a =10

So equation would be$\frac{y^2}{10^2} -\frac{x^2}{b^2} =1$

Since asymptotes are y =±5x/4

we have equation of both asymptotes is

$y^2-\frac{25x^2}{16} =0\\y^2/25-x^2/16 =0\\y^2/100-x^2/64 =0$

Since hyperbola will have equations same as asymptotes except with difference of constant terms as 1 instead of 0, we have

equation as

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