Question

How do you find all the critical points to graph `x^2 - y^2 + 9 = 0` including vertices, foci and asymptotes?

Answer 1

Center: `(0,0)`, vertices are at `(0,3) and (0,-3)`, focii are at`(0,3 sqrt 2) and (0,-3 sqrt 2)`, equation of asymptotes
of vertical hyperbola are `y=+-x`

Explanation:

`x^2-y^2+9=0 or y^2- x^2=9` or

`y^2/3^2- x^2/3^2=1`.The standard equation of vertical

hyperbola is `(y-k)^2/a^2-(x-h)^2/b^2=1; h,k` being center.

Here `h=0 ,k=0 , a=3 , b=3`. So center is at `(0,0)`

Vertices are `a=3` units from center . Therefore two vertices

are at `(0,3) and (0,-3)` , Focii are `c` units from center.

`c^2=a^2+b^2=9+9=18 :. c= 3 sqrt2`. Therefore focii are

`(0,3 sqrt 2) and (0,-3 sqrt 2)`. Hyperbola has two asymptotes

that intersect at the center of the hyperbola. The asymptotes

pass through the vertices of a rectangle of dimensions

`2 a=6 and 2 b=6` with its center at `(0,0) :.`

slope `+-a/b=+- 1`. Equation of asymptotes are

`y-0= +-1(x-0) or y=+-x`

graph{x^2-y^2+9=0 [-10, 10, -5, 5]}[Ans]