How do I find the equation of and graph an ellipse with the foci at (0,2) and (0,-2) and a major axis with a length of 8?

Question

This is a conics question for my college algebra class.

Answer 1

The distance between foci  knowing that the foci are over the y axis, the distance will be +2-(-2)=4 , the major axis is over the y axis also and then we have: c= distance from center to foci = 2;

a= half the semiaxis =8/2= 4; Now with the two distances known and by analysis the center of the ellipse is at the point (0,0) (origin), the equation may be expressed as ((x-h)^2/a^2) + ((y-k)^2/b^2) =1 h=0 and k =0 because the center is at the origin so x^2/a^2 + y^2/b^2 =1 => b^2*x^2 + a^2*y^2 = a^2*b^2 => 

If we define a=√C and b=√A, we have Ax^2 + Cy^2 +Dx + Ey =a^2*b^2. to obtain b we do:

c = √(a^2 - b^2) then b= 2√3; and then to get the equation in two forms we have:

Ax^2 + Cy^2 +0 +0= 12*16 = 12x^2 + 16y^2 - 192=0 => 4(3x^2 + 4y^2 - 48)=0

(3x^2 + 4y^2 - 48) this is the equation, the graph is as esy as to put y = and solve for x