How do you find the center, vertices, foci and eccentricity of #3x^2+4y^2-6x+16y+7=0#?
1 Answer
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
Now, the center is C( 1, -2 ).
Semi axes a = 2 and b =
Eccentricity
The major axis A'CA is along
major axis.
The vertices that are at distance a = 2, from the center
So, they are at
Likewise, the foci S and S" are at distance a e = 1, from C
So, they are at
I think that these data are sufficient to make a sketch of the
ellipse...