Question

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

Answer 1

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.