Conics Q (1 Viewer)

azureus88 · Jan 21, 2009

The diameter of ellipse x^2/a^2 + y^2/b^2 = 1 (a>b>0) through P(acos@, b sin@) meets circle x^2 + y^2 = a^2 at R(acos&,asin&). If the tangent to ellipse at P and tangent to the circle at R are concurrent with the right hand directrix of the ellipse, show that sec@ = 2/e.

not sure where to head with this question.

i found the tangent at P to be bx+aytan@=absec@ and tangent at R to be ax+bytan@=(a^2)(sec&). by substituting x=a/e and manipulation, i ended up with 1-e^2 = (esec& -1)/(esec@ -1).

Is there a way to simplify this or is there a better way around it?

vds700 · Jan 21, 2009

azureus88 said:
The diameter of ellipse x^2/a^2 + y^2/b^2 = 1 (a>b>0) through P(acos@, b sin@) meets circle x^2 + y^2 = a^2 at R(acos&,asin&). If the tangent to ellipse at P and tangent to the circle at R are concurrent with the right hand directrix of the ellipse, show that sec@ = 2/e.

not sure where to head with this question.

i found the tangent at P to be bx+aytan@=absec@ and tangent at R to be ax+bytan@=(a^2)(sec&). by substituting x=a/e and manipulation, i ended up with 1-e^2 = (esec& -1)(esec@ -1).

Is there a way to simplify this or is there a better way around it?

Im pretty siue theres a mistake with this question, it should be prove sec@ = 1/e

azureus88 · Jan 21, 2009

ok, so how did you come up with that answer?

vds700 · Jan 22, 2009

azureus88 said:
ok, so how did you come up with that answer?

I remember this question was posted before, and ppl said they got sec@ = 1/e.

As for doing the question, i cbf, someone smart like namu or trebla can help you

azureus88 · Jan 22, 2009

lol, k i'll wait then.

azureus88 · Jan 27, 2009

i dont think its sec@ = 1/e</SPAN> cause the demoninator would be zero.

Trebla · Jan 27, 2009

$$For symbol convenience let P be $ (a\cos\alpha,b\sin\alpha)$ and R be $(a\cos\beta,a\sin\beta)\\ $Equation of tangenet at P: (derive this) $\\bx\cos\alpha + ay\sin\alpha - ab = 0\\$Equation of tangent at R: (derive this)$\\x\cos\beta + y\sin\beta - a = 0\\\\$Now sub $x = \dfrac{a}{e}$ into both equations, giving$\\b\dfrac{a}{e}\cos\alpha + ay\sin\alpha - ab = 0\Rightarrow y=\dfrac{b\sec\alpha - \dfrac{b}{e}}{\tan\alpha}\\\\$and$\\ \dfrac{a}{e}\cos\beta + y\sin\beta - a = 0\Rightarrow y=\dfrac{a\sec\beta - \dfrac{a}{e}}{\tan\beta}$

$\\$Due to concurrency the y values are equal, so:$\\\dfrac{a\sec\beta - \dfrac{a}{e}}{\tan\beta} = \dfrac{b\sec\alpha - \dfrac{b}{e}}{\tan\alpha}\\\Rightarrow \dfrac{b(\sec\alpha - \dfrac{1}{e})}{a(\sec\beta - \dfrac{1}{e})} = \dfrac{\tan\alpha}{\tan\beta}\\$But the diameter passing through both points P and R must pass through the origin, so the gradients of OP and OR are equal, hence:$\\\dfrac{b\sin\alpha}{a\cos\alpha}=\dfrac{a\sin\beta}{a\cos\beta}\Rightarrow \tan\beta=\dfrac{b}{a}\tan\alpha\\\Rightarrow\dfrac{b(\sec\alpha - \dfrac{1}{e})}{a(\sec\beta - \dfrac{1}{e})} = \dfrac{a}{b}\\\Rightarrow\dfrac{b^2(\sec\alpha - \dfrac{1}{e})}{a^2(\sec\beta - \dfrac{1}{e})} = 1$
$\\$But $ b^2=a^2(1-e^2)\\\Rightarrow (1-e^2)(\sec\alpha - \dfrac{1}{e})=\sec\beta - \dfrac{1}{e}\\\Rightarrow e + (1-e^2)\sec\alpha = \sec\beta\\$Square both sides$\\e^2 + 2e(1-e^2)\sec\alpha + (1-e^2)^2\sec^2\alpha=\sec^2\beta\\\Rightarrow -(1 - e^2) + 2e(1-e^2)\sec\alpha + (1-e^2)^2\sec^2\alpha=\sec^2\beta - 1\\\Rightarrow (1 - e^2)(-1 + 2e\sec\alpha + (1-e^2)\sec^2\alpha)=\tan^2\beta\\\Rightarrow (1 - e^2)(-1 + 2e\sec\alpha + (1-e^2)\sec^2\alpha)=\dfrac{b^2}{a^2}\tan^2\alpha$

$$Again using $ b^2=a^2(1-e^2)\\\Rightarrow (1 - e^2)(-1 + 2e\sec\alpha + (1-e^2)\sec^2\alpha)=(1-e^2)\tan^2\alpha\\\Rightarrow 2e\sec\alpha + (1-e^2)\sec^2\alpha=\tan^2\alpha+1\\\Rightarrow 2e\sec\alpha + (1-e^2)\sec^2\alpha=\sec^2\alpha\\\Rightarrow 2e + (1-e^2)\sec\alpha=\sec\alpha\\\Rightarrow 2e =\sec\alpha(1-(1-e^2))\\\Rightarrow \sec\alpha = \dfrac{2}{e}$

azureus88 · Jan 27, 2009

omg i love you trebla, you're my new hero! thanks!

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