Question 35472
{{{(x+5)^2 + 4(y-4)^2 = 16}}}


Divide both sides of the equation by 16 in order to set it equal to 1, which is standard form for an ellipse.

{{{((x+5)^2)/16 + (4(y-4)^2)/16 = 16/16}}}

{{{ ((x+5)^2)/16 + ((y-4)^2)/4 = 1}}}


From this the center of the ellipse will be at (-5,4), the major radius will be in the x direction with {{{a^2=16}}} so a = 4, and the minor radius will be in the y direction with {{{b^2 = 4}}}, so b = 2.  Since for ellipses, {{{a^2 = b^2 + c^2}}}, {{{16 = 4 +c^2}}} so c = focal distance which is {{{c= sqrt(12) = 2*sqrt(3)}}}, applied in the x direction.


From that you should be able to sketch the graph and find the vertices, foci, and other information.


R^2 at SCC