# Thread: Complete the square.. and.

1. ## Complete the square.. and.

Okay I have to "little" problems to do like my teacher said .

1.) Complete the square to write the equation in standard form and sketch the graph.

x^2+4x+3y^2-5=0

2.) Determine an equation for an ellipse with center (5,2), semi major axis = 3, and one focus at (7,-2).

1.) Complete the square to write the equation in standard form and sketch the graph.

x^2+4x+3y^2-5=0

...
$x^2+4x+3y^2-5=0~\implies~x^2+4x+{\color{blue} 4} +3y^2=5+{\color{blue}4}$ Now divide both sides by 9:

$\frac{(x+2)^2}{9} + \frac{y^2}{3}=1$

That's the equation of an ellipse with center at C(-2, 0) and the semi-axes $a = 3$ and $b = \sqrt{3}$

...

2.) Determine an equation for an ellipse with center (5,2), semi major axis = 3, and one focus at (7,-2).
I assume that there is a typo:

You are supposed to know that the distance between the center and one focus of the ellipse is called the linear excentricity e. It is calculated by:

$e^2+b^2=a^2$

With your problem $e^2=20$ that means it is greater than a so a can't be the semi major axis.

The points C(5, 2) and F(7, -2) indicate an ellipse which has been rotated by 63.4° clockwise.

Before I do the calculations please check the wording of your problem.

4. Ooops.

The type was

(5,-2)

question again is.

Determine an equation for an ellipse with center (5,-2), semimajor axis = 3, and one focus at (7,2).

...

Determine an equation for an ellipse with center (5,-2), semimajor axis = 3, and one focus at (7,-2).
The distance between the center and the focus is 2, that means e = 2.

Therefore:

$2^2+b^2=3^2~\implies~b=\sqrt{5}$

Plug in all the values you know into the general equation of an ellipse:

$\frac{(x-5)^2}{3^2}+\frac{(y+2)^2}{(\sqrt{5})^2}=1 ~\implies~ \frac{(x-5)^2}{9}+\frac{(y+2)^2}{5}=1$