More ellipse problems

Jul 2009
338
14
Singapore
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 \(\displaystyle \sin\theta\)
 
Last edited:
Jun 2009
806
275
Check the coordinates of X, the center of the circle through P,T and N.

It cannot be
 
Jul 2009
338
14
Singapore
That is the answer given in the answer page.
 
Jul 2009
338
14
Singapore
yes, but i have found that it is correct, X is the mid-point of TN
 
Jun 2009
806
275
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?
 
Jul 2009
338
14
Singapore
\(\displaystyle a=3\sqrt{2}, b=3\) so the answers you got is the same are they not?
 
Jun 2009
806
275
\(\displaystyle a=3\sqrt{2}, b=3\) so the answers you got is the same are they not?
Co-ordinates of mid point of TN are

\(\displaystyle \frac{3\sqrt{2}}{2cosθ}, \frac{-3sinθ}{2} \)
 
Jun 2009
806
275
\(\displaystyle a=3\sqrt{2}, b=3\) so the answers you got is the same are they not?
Co-ordinates of mid point of TN are

\(\displaystyle (\frac{3\sqrt{2}}{2\cos\theta}, \frac{-3\sin\theta}{2})\)
 
Jul 2009
338
14
Singapore
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