More ellipse problems

**arze** · May 10th 2010, 05:06 PM

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 $\sin\theta$

**sa-ri-ga-ma** · May 11th 2010, 02:52 AM

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

It cannot be

**arze** · May 11th 2010, 03:58 PM

That is the answer given in the answer page.

**sa-ri-ga-ma** · May 11th 2010, 05:04 PM

Originally Posted by arze

That is the answer given in the answer page.

TPN is a right angled triangle. So the circumcenter must be the mid point of TN.

**arze** · May 11th 2010, 08:46 PM

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

**sa-ri-ga-ma** · May 11th 2010, 09:05 PM

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?

**arze** · May 11th 2010, 09:33 PM

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

**sa-ri-ga-ma** · May 12th 2010, 01:49 AM

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

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

**sa-ri-ga-ma** · May 12th 2010, 01:51 AM

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

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

**arze** · May 12th 2010, 01:54 AM

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

**sa-ri-ga-ma** · May 12th 2010, 02:32 AM

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

**arze** · May 12th 2010, 02:51 AM

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

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