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)?

1 Answer
Oct 4, 2016

Answer:

# 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#