Conics Question (1 Viewer)

kev-kun · Mar 16, 2014

P is a point on the ellipse x^2/a^2 + y^2/b^2 =1 with centre O. line drawn from O, parallel to the tangent to the ellipse at P, meets the ellipse at Q. Prove that the area of the triangle OPQ is independent of the position P.

Found this question quite tough :/ Any help is appreciated ^^

Trebla · Mar 16, 2014

kev-kun said:
P is a point on the ellipse x^2/a^2 + y^2/b^2 =1 with centre O. line drawn from O, parallel to the tangent to the ellipse at P, meets the ellipse at Q. Prove that the area of the triangle OPQ is independent of the position P.

Found this question quite tough :/ Any help is appreciated ^^

Let the points have the parametric coordinates $P(a\cos\alpha, b\sin\alpha)$ and $Q(a\cos\beta, b\sin\beta)$ .

I will use the perpendicular distance approach. The length of OP is given by

$OP = \sqrt{a^2\cos^2\alpha+b^2\sin^2\alpha}$

The equation of OP is given by

$y=\dfrac{b\sin\alpha}{a\cos\alpha} x$

Hence the perpendicular distance (denoted as d) between Q and OP is given by

$d=\dfrac{|\frac{b\sin\alpha}{a\cos\alpha} (a\cos\beta) - b\sin\beta|}{\sqrt{\frac{b^2\sin^2\alpha}{a^2\cos^2\alpha}+1}}\\\\=\dfrac{ab\sin\alpha\cos\beta-ab\cos\alpha\sin\beta}{\sqrt{a^2\cos^2\alpha+b^2 \sin ^2 \alpha}}\\\\=\dfrac{|ab\sin(\alpha-\beta)|}{\sqrt{a^2\cos^2\alpha+b^2\sin^2\alpha}}$

The area of the triangle (denoted as A) is therefore given by

$A=\dfrac{d}{2}OP\\\\=\dfrac{|ab\sin(\alpha-\beta)|}{2}$

BUT we haven't yet used the fact that OQ is parallel to the tangent at P. This gives the condition that

$-\dfrac{b\cos\alpha}{a\sin\alpha}=\dfrac{b\sin\beta}{a\cos\beta}\\\\\Rightarrow \cos(\alpha-\beta)=0\\\\\Rightarrow \sin^2(\alpha-\beta)=1\\\\\Rightarrow |\sin(\alpha-\beta)|=1$

Therefore

$A=\dfrac{|ab|}{2}$

which is independent of the positions of P (and Q)

kev-kun · Mar 17, 2014

Ohh thanks for the help ^^ But just one question what did you mean in:
BUT we haven't yet used the fact that OQ is parallel to the tangent at P. This gives the condition that
Not sure what you did there.

Trebla · Mar 17, 2014

kev-kun said:
Ohh thanks for the help ^^ But just one question what did you mean in:
BUT we haven't yet used the fact that OQ is parallel to the tangent at P. This gives the condition that
Not sure what you did there.

I just equated the gradients of the tangent and the line OQ

Conics Question (1 Viewer)

kev-kun

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Trebla

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kev-kun

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Trebla

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