Question

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

Answer 1

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

`(3(x-1)^2-3)+(4(y+2)^2-16)+7=0` Reducing to the standard form, it is

`(x-1)^2/2^2+(y+2)^2/(sqrt 3)^2=1`.

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

Semi axes a = 2 and b = `sqrt 3`..

Eccentricity `e = sqrt(a^2-b^2)/a=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( 1, - 2)`.

So, they are at `A( 3, -2) and A'(-1, -2)`.

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

So, they are at `S( 2, -2 ) and S'( 0, -2 )`.

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

ellipse...