# How do you find all the critical points to graph 25x^2- 4y^2=100 including vertices, foci and asymptotes?

Jan 1, 2017

The center is $\left(0 , 0\right)$
The vertices are$\left(2 , 0\right)$ and $\left(- 2 , 0\right)$
The equations of the asymptotes are $y = \frac{5}{2} x$ and $y = - \frac{5}{2} x$
The foci are $F = \left(\sqrt{29} , 0\right)$ and $F ' = \left(- \sqrt{29} , 0\right)$

#### Explanation:

Let's divide throughly by $100$

$\frac{25 {x}^{2}}{100} - \frac{4 {y}^{2}}{100} = \frac{100}{100}$

${x}^{2} / 4 - {y}^{2} / 25 = 1$

This is a hyperbola whose general equation is

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

The center is $\left(h , h\right) = \left(0 , 0\right)$

The vertices are $A = \left(h + a , k\right) = \left(2 , 0\right)$

and $A ' = \left(h - a , k\right) = \left(- 2 , 0\right)$

The asymptotes are obtained from the equation

${x}^{2} / 4 - {y}^{2} / 25 = 0$

$\left(\frac{x}{2} + \frac{y}{5}\right) \left(\frac{x}{2} - \frac{y}{5}\right) = 0$

The equations of the asymptotes are $y = \frac{5}{2} x$ and $y = - \frac{5}{2} x$

We calculate $c = \pm \sqrt{{a}^{2} + {b}^{2}} = \sqrt{4 + 25} = \sqrt{29}$

The foci are $F = \left(h + c , k\right) = \left(\sqrt{29} , 0\right)$ and

$F ' = \left(h - c , k\right) = \left(- \sqrt{29} , 0\right)$

graph{(x^2/4-y^2/25-1)(y-5/2x)(y+5/2x)=0 [-6.24, 6.244, -3.12, 3.12]}