# Identify Critical Points

## Key Questions

• $\left(h , k\right) \to \left(x , y\right)$ represents the center of the hyperbola, ellipse, and circle.

$\left(h , k\right) \to \left(x , y\right)$ represents the vertex of the parabola.

• If a hyperbola has an equation of the form $\frac{{x}^{2}}{{a}^{2}} - \frac{{y}^{2}}{{b}^{2}} = 1$ $\left(a > 0 , b > 0\right)$, then its slant asymptotes are $y = \pm \frac{b}{a} x$.

Let us look at some details.

By observing,

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

by subtracting $\frac{{x}^{2}}{{a}^{2}}$,

$R i g h t a r r o w - \frac{{y}^{2}}{{b}^{2}} = - \frac{{x}^{2}}{{a}^{2}} + 1$

by multiplying by $- {b}^{2}$,

$R i g h t a r r o w {y}^{2} = \frac{{b}^{2}}{{a}^{2}} {x}^{2} - {b}^{2}$

by taking the square-root,

$R i g h t a r r o w y = \pm \sqrt{\frac{{b}^{2}}{{a}^{2}} {x}^{2} - {b}^{2}} \approx \pm \sqrt{\frac{{b}^{2}}{{a}^{2}} {x}^{2}} = \pm \frac{b}{a} x$

(Note that when $x$ is large, $- {b}^{2}$ is negligible.)

Hence, its slant asymptotes are $y = \pm \frac{b}{a} x$.

• Critical points are points on the graph of a function where the first order derivative changes signs or equals to zero.

(Iam assuming you mean or you want something else )

• Make the equation be in the form

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

or

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

If $x$ is on front, the hyperbola opens horizontally
If $y$ is on front, the hyperbola opens vertically