Question

I need some help with this question on conics:
Find an equation for the ellipse with the foci (1,1) and (-1,-1) and a major axes of length 4.

Answer 1

This is a shifted conic, i need to find the equation of this shifted conic

Answer 2

this looks like an ellipse with a slanted major axis, is that correct?

Answer 3

yes thats correct

Answer 4

It lies on the rotated XY coordinate axes

Answer 5

ok the axis is the line y=x, since it goes grough both foci points
2a = 4
a = 2
imagine a triangle with equal legs and hypotenuse of 2
x =y = 2/sqrt2 = sqrt2
verticis are (sqrt2, sqrt2), (-sqrt2, -sqrt2)

Answer 6

minor axis must be perpendicular so on the line y=-x
length of minor axis is 2b, c is length of foci point from origin --> c=sqrt2
a^2 = b^2 +c^2
2^2 = b^2 +(sqrt2)^2
4 = b^2 +2
b = sqrt2

again imagine 45-45 -90 triangle with hypotenuse of sqrt2
x=y=1
minor verticis = (-1,1) (1,-1)

Answer 7

Shoudlnt we incorporate those formulas x=Xcos(theta)-Ysin(theta)
and y=Xsin(theta)+Ycos(theta)

Answer 8

because i am looking for the equation of the rotated ellipse

Answer 9

yes to get the equation we have to rotate 45 degrees to get major axis on line y=x
now we have a,b
so equation of non-rotated ellipse is:
\[\frac{x^{2}}{4}+\frac{y^{2}}{2} = 1\]
theta = 45
sin(45) = cos(45) = sqrt2/2
\[x' = \sqrt{2}/2(x-y)\]
\[y' = \sqrt{2}/2 (x+y)\]
replace x with x' and y with y' in equation

Answer 10

Okay i got the equation of the non rotated ellipse, but now i understand how to get the equation of the rotated ellipse. Thank you.

Answer 11

if you could take a look at my other question regarding a helix and curvature, it would be much appreaciated

Answer 12

ok