Find the equation of ellipse whose center Is the origin one focus at (2,2) and length of the semi minor axis is √8

Answer 1

since the centre is (0,0) and the major axis passes through (2,2), you are
looking at an ellipse which has been rotated 45° from its standard position.
So in that standard position, c = √(2^2 + 2^2) = √8
but the semi minor axis is also √8, and would not change for any rotation
so we have a^2 = b^2 + c^2 = 8+8 = 16 , (from b = √8)
and a = 4
so the ellipse before its rotation was x^2/16 + y^2/8 = 1
8x^2 + 16y^2 = 128

Now apply the rotation matrix to this equation:
x = u cosθ + vsinθ = u√2/2 + v√2/2 , since θ = 45°
y = -usinθ + vcosθ = -u√2/2 + v√2/2
where u and v will be the x and y of our rotated equation

8(u√2/2 + v√2/2)^2 + 16(-u√2/2 + v√2/2)^2 = 128
8(u^2/2 + uv + v^2/2) + 16(u^2/2 - uv + v^2/2) = 128
4u^2 + 8uv + 4v^2 + 8u^2 - 16uv + 8v^2 = 128
12u^2 - 8uv + 12v^2 = 128
3u^2 - 2uv + 3v^2 = 32

or, in terms of x and y:
3x^2 - 2xy + 3y^2 = 32

whewww!, I graphed it on Desmos and it is correct, yeahhh

Answer 2

so you have
c = 2√2
b = 2√2
so a = 4
If the major axis were along the x-axis, then the equation would be
x^2/16 + y^2/8 = 1
But you have rotated it by 45° so you need to apply the rotation matrix to obtain the new x' and y' equation