# What are the critical points of x^2/9 + y^2/36 = 1?

Nov 10, 2016

The axes are $= 12 \mathmr{and} 6$
The center is $\left(0 , 0\right)$
The foci are F$= \left(0 , 5\right)$ and $\left(0 , - 5\right)$

#### Explanation:

This is the equation of an ellipse.
The major axis $= 12$
The minor axis $= 6$

${\left(x - 0\right)}^{2} / {3}^{2} + {\left(y - 0\right)}^{2} / {6}^{2} = 1$
The center is $\left(0 , 0\right)$

$c = \sqrt{36 - 9} = \sqrt{25} = 5$

The foci are F$= \left(0 , 5\right)$ and $\left(0 , - 5\right)$

graph{(x^2/9)+(y^2/36)=1 [-14.24, 14.23, -7.12, 7.12]}

Nov 10, 2016

Critical points are $\left(0 , 6\right)$, $\left(0 , - 6\right)$, $\left(3 , 0\right)$ and $\left(- 3 , 0\right)$.

#### Explanation:

A critical point is a point on a curve, where either the derivative is not defined or its value is zero.

As ${x}^{2} / 9 + {y}^{2} / 36 = 1$, we have

$\frac{2 x}{9} + \frac{2 y}{36} \times \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x}{9} \times \frac{36}{2 y} = - \frac{4 x}{y}$,

which is zero when $x = 0$ ($y$-axis) and is not defined when $y = 0$ ($x$-axis).

Note that when $x = 0$, $y = \pm 6$ and when $y = 0$, $x \pm 3$, i.e.

Critical points are $\left(0 , 6\right)$, $\left(0 , - 6\right)$, $\left(3 , 0\right)$ and $\left(- 3 , 0\right)$.

In fact, it is an ellipse with major axis at $y$-axis and minor axis at $x$-axis.
graph{x^2/9+y^2/36=1 [-20, 20, -10, 10]}