it's late and i cant get these 2 question to fall out! help please :)

I have point $\displaystyle p(3\sqrt2 cos \theta,3sin \theta)$ on ellipse

$\displaystyle

\frac{x^2}{18}+\frac{y^2}{9}=1

$

I have the tangent and normal to this ellipse meet Y-axis at T and N. Then i found the centre of the circle through T,N,P to be

$\displaystyle

X=(0,\frac{3}{2}[\frac{1-sin^2\theta}{sin\theta}]

$

The last part asks for the equation of the locus of Q on PX produced such that X is the midpoint of PQ

so ive let Q=(x,y) then

$\displaystyle

\frac{x+3\sqrt(2)}{2}=0

$

$\displaystyle

\frac{y+3sin\theta}{2}=\frac{3}{2}[\frac{1-sin^2\theta}{sin\theta}]

$

so $\displaystyle Q=(-3\sqrt(2),3[\frac{1-2sin^2\theta}{sin\theta}])$

but i cant find the locus. The book says

$\displaystyle

\frac{2x^2}{9}+\frac{9}{16y^2}=1

$

i cant get this with my Q. Maybe im making a mistake or my Q is wrong or the book is wrong......

2)

Also having trouble with this:

the tangent at $\displaystyle P(a cos\theta,bsin\theta)$ to

ellipse $\displaystyle b^2x^2+a^2y^2=a^2b^2$

cuts the Y-axis at Q. The normal at P is parallell to the line joining Q to one focus S'. If S is the other focus, show that PS is parallel to the y-axis.

found Q to be $\displaystyle (0,\frac{b}{sin\theta}) $

focus $\displaystyle S'=(ae,0)$ $\displaystyle S=(-ae,0) $ with $\displaystyle a^2-b^2=a^2e^2$

grad $\displaystyle QS=\frac{\frac{b}{sin\theta}}{-ae}=\frac{-b}{aesin\theta}$

but QS parallel to normal so grad normal= grad QS

so

$\displaystyle

\frac{asin\theta}{bcos\theta}=\frac{-b}{aesin\theta}

$

and grad $\displaystyle PS=\frac{bsin\theta}{acos\theta + ae}$

now im assuming i use the grad QS=grad normal to show the denominator of grad PS=0 but i cant seem to get it....