What are the critical points of $\frac{x^{2}}{9} + \frac{y^{2}}{36} = 1$ $x^2/9 + y^2/36 = 1$ ?

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Answer 1 · 2016-11-10T07:29:32

The axes are $= 12 and 6$ $=12 and 6$
The center is $(0, 0)$ $(0,0)$
The foci are F $= (0, 5)$ $=(0,5)$ and $(0, - 5)$ $(0,-5)$

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

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

$c = \sqrt{36 - 9} = \sqrt{25} = 5$ $c=sqrt(36-9)=sqrt25=5$

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

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

Answer 2 · 2016-11-10T07:30:18

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

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

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

$\frac{2 x}{9} + \frac{2 y}{36} \times \frac{d y}{d x} = 0$ $(2x)/9+(2y)/36xx(dy)/(dx)=0$

or $\frac{d y}{d x} = - \frac{2 x}{9} \times \frac{36}{2 y} = - \frac{4 x}{y}$ $(dy)/(dx)=-(2x)/9xx36/(2y)=-(4x)/y$ ,

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

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

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

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

What are the critical points of x29+y236=1x^2/9 + y^2/36 = 1?