# Question f1260

Dec 2, 2015

$y = \pm \frac{5}{4} {\left(16 - {x}^{2}\right)}^{0.5}$

#### Explanation:

$x = 4 \cos \theta$ & $y = 5 \sin \theta$
Square both sides
${x}^{2} = 16 {\cos}^{2} \theta$ & ${y}^{2} = 25 {\sin}^{2} \theta$
Notice that ${\sin}^{2} \theta = 1 - {\cos}^{2} \theta$
${y}^{2} = 25 \left(1 - {\cos}^{2} \theta\right) \to$ Equation 1

Substitute ${\cos}^{2} \theta = {x}^{2} / 16$ into Equation 1.

${y}^{2} = 25 \left(1 - {x}^{2} / 16\right)$
Rearrange this to get:
$y = \pm \frac{5}{4} {\left(16 - {x}^{2}\right)}^{0.5}$

From your domain $- \frac{\pi}{2} \le \theta \le \frac{\pi}{2}$

We find that x ranges from {x;0<=x,+4}
and y ranges from {y;-5<=y<=5}#.
The fact that $x = 4 \cos \theta$ & $y = 5 \sin \theta$ takes the form of a parametric curve (oval shape), we can draw an oval that is restricted by the given conditions.
graph{5/4(16-x^2)^0.5 [0, 4, -5, 5.5]}
The graph also reflects on the x-axis.
graph{-5/4(16-x^2)^0.5 [0, 4, -5, 5.5]}
Put the 2 graphs together.

The arrow should be pointing clockwise.