Question

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

Answer 1

Answer:

d. ellipse; $23\degree$

Step-by-step explanation:

The given equation is;

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

Comparing to the general equation;

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

$A=9,B=4,C=5$

$\cot(2\theta)=\frac{9-5}{4}$

$\cot(2\theta)=\frac{4}{4}$

$\cot(2\theta)=1$

$2\theta=\cot^{-1}$

$2\theta=45\degree$

$\theta=22.5\degree$

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.