The tangent and normal at \(\displaystyle P(3\sqrt{2}\cos\theta, 3\sin\theta)\) to the ellipse \(\displaystyle \frac{x^2}{18}+\frac{y^2}{9}=1\)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. \(\displaystyle X(0,\frac{3}{2}(\frac{1}{\sin\theta}-\sin\theta))\)

I let Q(x,y)

so \(\displaystyle (0,\frac{3}{2}(\frac{1}{\sin\theta}-\sin\theta))=(\frac{3\sqrt{2}\cos\theta+x}{2},\frac{3\sin\theta+y}{2})\)

so \(\displaystyle 0=\frac{3\sqrt{2}\cos\theta+x}{2}\)---1

and \(\displaystyle \frac{3}{2}(\frac{1}{\sin\theta}-\sin\theta)=\frac{3\sin\theta+y}{2}\)---2

I worked out 1 and got:

\(\displaystyle \cos\theta=\frac{\sqrt{2}x}{6}\)

But i cannot express 2 in terms of \(\displaystyle \sin\theta\)

I have completed all but the last part of this question and got right answers. \(\displaystyle X(0,\frac{3}{2}(\frac{1}{\sin\theta}-\sin\theta))\)

I let Q(x,y)

so \(\displaystyle (0,\frac{3}{2}(\frac{1}{\sin\theta}-\sin\theta))=(\frac{3\sqrt{2}\cos\theta+x}{2},\frac{3\sin\theta+y}{2})\)

so \(\displaystyle 0=\frac{3\sqrt{2}\cos\theta+x}{2}\)---1

and \(\displaystyle \frac{3}{2}(\frac{1}{\sin\theta}-\sin\theta)=\frac{3\sin\theta+y}{2}\)---2

I worked out 1 and got:

\(\displaystyle \cos\theta=\frac{\sqrt{2}x}{6}\)

But i cannot express 2 in terms of \(\displaystyle \sin\theta\)

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