# How do you find all the critical points to graph 36x^2 - 100y^2 - 72x + 400y = 3964 including vertices, foci and asymptotes?

Sep 28, 2017

Vertices are $\left(11 , 2\right)$ and $\left(- 9 , 2\right)$, foci are (1±2sqrt(34),2) and asymptotes are $y = \frac{3}{5} x + \frac{7}{5}$ and $y = - \frac{3}{5} x + \frac{13}{5}$.

#### Explanation:

The graph is for a hyperbola.
The standard form of equation for a hyperbola is
x^2/(a^2) - y^2/(b^2) = ±1. $\left(a > 0 , b > 0\right)$
What we have to do is to deform the fomula to this form.

[step1] Let's complete the square.
$36 {x}^{2} - 100 {y}^{2} - 72 x + 400 y = 3964$
$36 \left({x}^{2} - 2 x\right) - 100 \left({y}^{2} - 4 y\right) = 3964$
36(x-1)^2-100(y-2)^2=3964+36×1-100×4
$36 {\left(x - 1\right)}^{2} - 100 {\left(y - 2\right)}^{2} = 3600$

[step2] divide both side by $3600$ to make a $1$ on the right side.
${\left(x - 1\right)}^{2} / \left({10}^{2}\right) - {\left(y - 2\right)}^{2} / \left({6}^{2}\right) = 1$

The hyperbola has vertices and foci on on the line $y = 2$.
Substituting $y = 2$ to the fomula yields $x = 11 , - 9$
Therefore, the vertices are $\left(11 , 2\right)$ and $\left(- 9 , 2\right)$.

The foci of ${x}^{2} / \left({a}^{2}\right) - {y}^{2} / \left({b}^{2}\right) = 1$ are (±sqrt(a^2+b^2),0) .
If the fomula were ${x}^{2} / \left({10}^{2}\right) - {y}^{2} / \left({6}^{2}\right) = 1$, the foci would be
(±sqrt(6^2+10^2),0) = (±2sqrt(34),0). Translate this $1$ to the right and $2$ to the up and the foci are (1±2sqrt(34),2).

The asymptopes of ${x}^{2} / \left({a}^{2}\right) - {y}^{2} / \left({b}^{2}\right) = 1$ are y=±(a/b)x and the asymptopes of ${\left(x - 1\right)}^{2} / \left({10}^{2}\right) - {\left(y - 2\right)}^{2} / \left({6}^{2}\right) = 1$ are therefore
y-2 = ±(6/10)(x-1)
$y = \frac{3}{5} x + \frac{7}{5}$ and $y = - \frac{3}{5} x + \frac{13}{5}$.