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