Join in on our 'Guess the Rep' Competition! >>

HELP....FP2 question

Can someone please help with me this question

Show that S(√3,0) 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.

I can do the first bit but not the second bit.

Reply 1

geo106

write in the form

x^2/a^2 + y^2/b^2=1

ie

x^2/12 + y^2/9 =1

use b^2=a^2(1-e^2)
to find e

then focus = ae

it shoud work out

Reply 2

AHW

yeah...I know how to do this part but i can't solve the second part for the locus of Q.

Reply 3

steve10

You've got 3x² + 4y² = 36.

do implicit differentation to get y' = -3x/4y.

Let the point P be (h,k).

Use y' and P(h,k) to get the slope of the tangent at P in terms of h and k.

Hence get the slope of the perpindicular to the tangent at P.
This is the slope of OQ.

OQ is a line passing through the origin.
Get the eqn of that line.

You have the coords of S and P.
get the eqn of the line passing through S and P.

Using these two line eqns ,solve to find their point of intersection, Q.
Call it Q(X,Y).

I got X = 3√3h/{3h + 4√3 - 4h} and Y = 4√3k/{3h + 4√3 - 4h}.
(haven't checked these values properly)

You can re-write the above coords as,

X = 3h*K, Y = 4k*K

Now let H² = (3h)² + (4k)².

Then,

X = Rcos@, Y = Rsin@ (a circle!)

Where R = KH and sin@ = 4k/H, cos@ = 3h/H.

Thus the coords of Q(X,Y) are defined by a circle.