Question

How do you find the vertices, asymptote, foci and graph `x^2-9y^2=25`?

Answer 1

Please see the explanation.

Explanation:

Here is a helpful Reference

The standard form for the Horizontal Transverse Axis type is:

`(x - h)^2/a^2 - (y - k)^2/b^2 = 1`

center: `(h, k)`

vertices: `(h -a, k) and (h + a, k)`

foci: `(h - sqrt(a^2 + b^2), k) and (h + sqrt(a^2 + b^2), k)`

Asymptotes:

`y = -b/a(x - h) + k and y = b/a(x - h) + k`

Given: `x^2 - 9y^2 = 25`

Put the given equation in standard form. Divide both sides by `25`:

`x^2/25 - 9y^2/25 = 1`

Flip the 9 down under the 25:

`x^2/25 - y^2/(25/9) = 1`

Write the denominators as squares:

`x^2/5^2 - y^2/(5/3)^2 = 1`

Insert - 0s in the numerators:

`(x - 0)^2/5^2 - (y - 0)^2/(5/3)^2 = 1`

center: `(0, 0)`

vertices: `(-5, 0) and (5, 0)`

foci: `(-5sqrt(10)/3, 0) and (5sqrt(10)/3, 0)`

Asymptotes:

`y = -1/3x and y = 1/3x`