Question

How do you write an equation of an ellipse in standard form given minor axis has length 12 and its focal points are at (1, 2) and (4,-2)?

Answer 1

`592 (x-5) x + 24 x y - 60 y + 585 y^2-82844=0`

Explanation:

How do you write an equation of an ellipse in standard form given minor axis has length `12` and its focal points are at `f_1=(1, 2)` and `f_2=(4,-2)`?

The ellipse is slanted by a rotation of

`theta = arctan(2+2,1-4)`

it's center is at `p_0={x_0,y_0} = (f_1+f_2)/2` so deffining a rotation matrix

`R=((costheta,-sintheta),(sintheta,costheta))` so

`R=((4/5,3/5),(-3/5,4/5))`

`M=((1/a^2,0),(0,1/b^2))`

and calling `p = (x,y)` we have

`(p-p_0)R^Tcdot M cdot R (p-p_0) = 1`

as the generic ellipse equation, with certer in `p_0` and main axes `a, b`.

`c = norm(f_1-f_2)/2`

`a^2 = b^2+c^2`

Here `b = 12` and `c = 5/2` so `a = sqrt(12^2+(5/2)^2) = sqrt[601]/2`

The final equation is

`592 (x-5) x + 24 x y - 60 y + 585 y^2-82844=0`