Heinemann Pure 5 Exercise 4B Q.16

Having completed 4B, this is the only question to prove troublesome.

Show that S(rt3,0) is a focus of the ellipse with equation 3x^2 + 4y^2 = 36.
The origin is O and P is any point on the ellipse. A line is drawn from O perpendicular to the tangent to the ellipse at P and this line meets the line SP, produced if necessary, in the point Q. Show that the locus of Q is a circle.

The first part, proving the focus, is simple; it is the second part that bothers me. I have an equation for the perpendicular and one for SP. However, making them equal to get co-ordinates/equations for Q is proving rather messy...

All help welcome
Thanks!

Reply 1

dvs

The equation of the normal at P through the origin is:
y = [(sqrt[12]sint)/(3cost)] x (1)

And the equation of SP is:
y = [(sqrt[3] sint)/(2 cost - 1)] (x - sqrt[3]) (2)

(1) = (2):
[(sqrt[12]sint)/(3cost)] x = [(sqrt[3] sint)/(2 cost - 1)] (x - sqrt[3])
(2sqrt[3])sint . (2 cost - 1) . x = (sqrt[3] sint) . (3 cost) . (x - sqrt[3])
(4sqrt[3]sintcost - 2sqrt[3]sint)x = (3sqrt[3] sintcost)x - 9sintcost
(sqrt[3]cost - 2sqrt[3])x = -9cost
x = 3sqrt[3]cost/(2 - cost)

Plug this in (1):
y = (2sqrt[3]sint) . sqrt[3]/(2 - cost) = 6sint/(2 - cost)

And thus we have the coordinates of Q. Now:
x = 3sqrt[3]cost/(2 - cost)
=> cost = 2x/(x + 3sqrt[3])

y= 6sint/(2 - cost)
=> sint = sqrt[3]y/(x + 3sqrt[3])

Using the identity sin²t + cos²t = 1, we get:
3y² + 4x² = (x + 3sqrt[3])² = x² + 6sqrt[3]x + 27
3y² + 3x² - 6sqrt[3]x - 27 = 0
y² + x² - 2sqrt[3]x - 9 = 0
(x - sqrt[3])² + y² = 12

So the locus of Q is a circle center (sqrt[3], 0) with radius 2sqrt[3].