# How do you find the center, vertices, foci and eccentricity of 3x^2+4y^2-6x+16y+7=0?

Sep 23, 2016

See explanation.

#### Explanation:

Note the absence of xy-term. The product of the coefficients of x^2

and y^2 = 1`2 > 0. So, the equation represents an ellipse.

Reorganized, it becomes

$\left(3 {\left(x - 1\right)}^{2} - 3\right) + \left(4 {\left(y + 2\right)}^{2} - 16\right) + 7 = 0$ Reducing to the standard form, it is

${\left(x - 1\right)}^{2} / {2}^{2} + {\left(y + 2\right)}^{2} / {\left(\sqrt{3}\right)}^{2} = 1$.

Now, the center is C( 1, -2 ).

Semi axes a = 2 and b = $\sqrt{3}$..

Eccentricity $e = \frac{\sqrt{{a}^{2} - {b}^{2}}}{a} = \frac{1}{2}$

The major axis A'CA is along $y = - 2$ Vertices and foci lie on the

major axis.

The vertices that are at distance a = 2, from the center $C \left(1 , - 2\right)$.

So, they are at $A \left(3 , - 2\right) \mathmr{and} A ' \left(- 1 , - 2\right)$.

Likewise, the foci S and S" are at distance a e = 1, from C

So, they are at $S \left(2 , - 2\right) \mathmr{and} S ' \left(0 , - 2\right)$.

I think that these data are sufficient to make a sketch of the

ellipse...