Let P be any point on the conic r = 12/(3-sin x). Let F¹ and F² be the points (0, 0°) and (3, 90°) respectively. Show that PF¹ and PF² = 9 ?

1 Answer
May 6, 2018

#r = 12/{3-sin theta }#

We're asked to show #|PF_1| + |PF_2| =9 #, i.e. #P# sweeps out an ellipse with foci #F_1# and #F_2.# See the proof below.

Explanation:

Let's fix what I'll guess is a typo and say #P(r, theta)# satisfies

#r = 12/{3-sin theta }#

The range of sine is #pm 1# so we conclude #4 le r le 6.#

#3r - r sin theta = 12 #

# |PF_1| = |P - 0| = r #

In rectangular coordinates, #P=(r cos theta, r sin theta)# and #F_2 = (3 cos 90^circ, 3 sin 90^circ) = (0,3)#

#|PF_2|^2=|P-F_2|^2 = r^2 cos^2 theta + (r sin theta - 3)^3 #

#|PF_2|^2 = r^2 cos^2 theta + r^2 sin ^2 theta - 6 r sin theta + 9 #

#|PF_2|^2 = r^2 - 6 r sin theta + 9 #

#r sin theta = 3r -12#

#|PF_2|^2 = r^2 - 6 (3r - 12) + 9 #

#|PF_2|^2 = r^2 - 18r + 81 = (r-9)^2#

#|PF_2|=|r-9|#

#|PF_2|=9-r quad # since we already know #4 le r le 6.#

#|PF_1| + |PF_2| = r + 9 -r = 9 quad sqrt#