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The equations of various ellipse constructions

  • Thread starter Appleton
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Homework Statement


[/B]
The tangent and the normal at a point P([itex]3\sqrt2\cos\theta,3\sin\theta)[/itex]) on the ellipse [itex]\frac{x^2}{18}+\frac{y^2}{9}=1[/itex] meet the y axis at T and N respectively. If O is the origin, prove that [itex]OT.TN[/itex] 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


[/B]
The tangent of the ellipse is:

[itex]3\sqrt2ysin\theta+3xcos\theta=9\sqrt2[/itex]

When x=0, [itex]y=\frac{3}{sin\theta}=T[/itex]

The normal of the ellipse is:

[itex]3\sqrt2xsin\theta-3ycos\theta=9sin\theta\cos\theta[/itex]

When x=0, [itex] y=-3sin\theta=N[/itex]

[itex]OT.ON =(\frac{3}{sin\theta})(-3sin\theta) cos\pi[/itex]

[itex]OT.ON =9[/itex] and is therefore independent of the position P.

If X(x, y) is the centre of the circle through P, T and N, then

[itex]XT^2=XN^2\implies[/itex]

[itex]x^2+y^2-\frac{6y}{sin\theta}+\frac{9}{sin^2\theta}=x^2+y^2+6y\sin\theta+9sin^2\theta[/itex]

[itex]y=\frac{3}{2}(\frac{cos^2(\theta)}{sin\theta})[/itex]

[itex]XT^2=XP^2\implies[/itex]

[itex]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)[/itex]

[itex]x=0[/itex]

So X, the centre of the circle passing through P, T and N is [itex](0,\frac{3}{2}\frac{cos^2\theta}{sin\theta})[/itex]

If X is the midpoint of PQ

[itex]x_Q=2x_X-x_P[/itex]

[itex]x_X=0\implies x_Q=-x_P=-3\sqrt2cos\theta[/itex]

[itex]y_Q=2y_X-y_P[/itex]

[itex]y_X=\frac{3}{2}\frac{cos^2\theta}{sin\theta}\implies y_Q=\frac{3}{sin\theta}-6sin\theta[/itex]

[itex]\theta=\arccos(-\frac{x_X}{3\sqrt2})[/itex]

[itex]y=\frac{9\sqrt2}{\sqrt(18-x^2)}-2\sqrt(18-x^2)[/itex]

However my book gives a different answer:

[itex]\frac{2x^2}{9}+\frac{9}{16y^2}=1[/itex]

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.
 

Answers and Replies

  • #2
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Hi, I'm hoping no news is good news. It's the last part of the question that i'm most curious about. Is my book correct? Or am I correct? Or neither?
 

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