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Appleton
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Homework Statement
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The tangent and the normal at a point P(3\sqrt2\cos\theta,3\sin\theta)) on 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 that OT.TN is independent of the position 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 midpoint of PQ.
Homework Equations
The Attempt at a Solution
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The tangent of the ellipse is:
3\sqrt2ysin\theta+3xcos\theta=9\sqrt2
When x=0, y=\frac{3}{sin\theta}=T
The normal of the ellipse is:
3\sqrt2xsin\theta-3ycos\theta=9sin\theta\cos\theta
When x=0, y=-3sin\theta=N
OT.ON =(\frac{3}{sin\theta})(-3sin\theta) cos\pi
OT.ON =9 and is therefore independent of the position P.
If X(x, y) is the centre of the circle through P, T and N, then
XT^2=XN^2\implies
x^2+y^2-\frac{6y}{sin\theta}+\frac{9}{sin^2\theta}=x^2+y^2+6y\sin\theta+9sin^2\theta
y=\frac{3}{2}(\frac{cos^2(\theta)}{sin\theta})
XT^2=XP^2\implies
x^2+y^2-\frac{6y}{sin\theta}+\frac{9}{sin^2\theta}=x^2+y^2+6cos\theta(3cos\theta-\sqrt2x)+3sin\theta(3sin\theta-2y)
x=0
So X, the centre of the circle passing through P, T and N is (0,\frac{3}{2}\frac{cos^2\theta}{sin\theta})
If X is the midpoint of PQ
x_Q=2x_X-x_P
x_X=0\implies x_Q=-x_P=-3\sqrt2cos\theta
y_Q=2y_X-y_P
y_X=\frac{3}{2}\frac{cos^2\theta}{sin\theta}\implies y_Q=\frac{3}{sin\theta}-6sin\theta
\theta=\arccos(-\frac{x_X}{3\sqrt2})
y=\frac{9\sqrt2}{\sqrt(18-x^2)}-2\sqrt(18-x^2)
However my book gives a different answer:
\frac{2x^2}{9}+\frac{9}{16y^2}=1
I've graphed both equations and my solution seems to correlate more closely to what I would expect than the book's solution. I would very much appreciate it if someone is able to identify any errors or more efficient methods that I may have missed.