Q4: Identify the graph of the equation and and find theta zero to the nearest degree.

Question

asked Jun 25, 2020 59.0k views

1 Answer

Alireza Farahani · Answer 1 · 2020-07-01T12:47:24+0000

Answer:

d. ellipse;
$23\degree$

Explanation:

The given equation is;

9x^2+4xy + 5y^2-40=0 .

Comparing to the general equation;

Ax^2+Bxy+Cy^2+Dx+Ey+F=0

A=9,B=4,C=5

$\cot(2\theta)=(9-5)/(4)$

$\cot(2\theta)=(4)/(4)$

$\cot(2\theta)=1$

$2\theta=\cot^(-1)$

$2\theta=45\degree$

$\theta=22.5\degree$

To the nearest degree is
$23\degree$

Since the coefficient of the
x^2 is not equal to the coefficient of
y^2 but the signs are the same, the conic is an ellipse.