1. ## More ellipse problems

The tangent and normal at $P(3\sqrt{2}\cos\theta, 3\sin\theta)$ to the ellipse $\frac{x^2}{18}+\frac{y^2}{9}=1$meet the y axis at T and N respectively. If O is the origin, prove the OT.ON is independent of P. Find the coordinates of X, the centre of the circle through P,T and N. Find also the equation of the locus of the point Q on PX produced such that X is the mid-point of PQ.
I have completed all but the last part of this question and got right answers. $X(0,\frac{3}{2}(\frac{1}{\sin\theta}-\sin\theta))$
I let Q(x,y)
so $(0,\frac{3}{2}(\frac{1}{\sin\theta}-\sin\theta))=(\frac{3\sqrt{2}\cos\theta+x}{2},\fra c{3\sin\theta+y}{2})$
so $0=\frac{3\sqrt{2}\cos\theta+x}{2}$---1
and $\frac{3}{2}(\frac{1}{\sin\theta}-\sin\theta)=\frac{3\sin\theta+y}{2}$---2
I worked out 1 and got:
$\cos\theta=\frac{\sqrt{2}x}{6}$
But i cannot express 2 in terms of $\sin\theta$

2. Check the coordinates of X, the center of the circle through P,T and N.

It cannot be

4. Originally Posted by arze
TPN is a right angled triangle. So the circumcenter must be the mid point of TN.

5. yes, but i have found that it is correct, X is the mid-point of TN

6. Originally Posted by arze
yes, but i have found that it is correct, X is the mid-point of TN
Equation of the tangent at P is

xcosθ/a + ysinθ/b = 1

It meets the x-axis at T (asecθ, 0).

Equation of the normal is

axsec(θ) - bycosec(θ) = a^2 - b^2

It meets the y-axis at N (0, [b^2 - a^2]/bcsc(θ))

I am getting the mid point of TN something different. Why?

7. $a=3\sqrt{2}, b=3$ so the answers you got is the same are they not?

8. Originally Posted by arze
$a=3\sqrt{2}, b=3$ so the answers you got is the same are they not?
Co-ordinates of mid point of TN are

$\frac{3\sqrt{2}}{2cosθ}, \frac{-3sinθ}{2}$

9. Originally Posted by arze
$a=3\sqrt{2}, b=3$ so the answers you got is the same are they not?
Co-ordinates of mid point of TN are

$(\frac{3\sqrt{2}}{2\cos\theta}, \frac{-3\sin\theta}{2})$

10. sorry, i forgot to tell you that T and N are both on the y-axis, there was i typo in my original post, i have corrected it already, there is no need to find the point of the x-axis intersect. very sorry

11. Originally Posted by arze
sorry, i forgot to tell you that T and N are both on the y-axis, there was i typo in my original post, i have corrected it already, there is no need to find the point of the x-axis intersect. very sorry
--

so y = 3/sinθ - 6sinθ and x = - 3sqrt(2) cosθ

12. i use the $\sin^2\theta+\cos^2\theta\equiv 1$ formula.
but i still can't get the answer which is $\frac{2x^2}{9}+\frac{9}{16y^2}=1$