Hyperbola Locus (2 Viewers)

wrumble12 · Dec 20, 2016

If R is the point (2(cos∅ - sin∅), 3(cos∅ + sin∅)), show that the locus of R is the ellipse x^2/18 + y^2/8 =1

Thankyou!

InteGrand · Dec 20, 2016

wrumble12 said:
If R is the point (2(cos∅ - sin∅), 3(cos∅ + sin∅)), show that the locus of R is the ellipse x^2/18 + y^2/8 =1

Thankyou!

$\noindent Observe that $\cos \phi -\sin \phi \equiv \sqrt{2}\cos \left(\phi +\frac{\pi}{4}\right)$ and $\cos \phi + \sin \phi \equiv \sqrt{2}\cos\left(\phi -\frac{\pi}{4}\right) \equiv \sqrt{2}\sin\left(\phi + \frac{\pi}{4}\right)$. So $R$'s coordinates satisfy$

$\begin{align*}x = 2\sqrt{2} \cos \left(\phi +\frac{\pi}{4}\right), y = 3\sqrt{2}\sin \left(\phi +\frac{\pi}{4}\right).\end{align*}$

$\noindent Substitute $\theta = \phi +\frac{\pi}{4}$, then $R$'s coordinates satisfy $x = 2\sqrt{2}\cos \theta, y = 3\sqrt{2}\sin \theta$. As we know, these are parametric equations for the ellipse $\frac{x^{2}}{8} + \frac{y^{2}}{18} = 1$, since $\left(2\sqrt{2}\right)^{2} = 8$ and $\left(3\sqrt{2}\right)^{2} = 18$ $\Big{(}$that $x = a\cos \theta, y = b\sin \theta$ are parametric equations for the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1$ follows from Pythagorean trig. identities.$\Big{)}$ (Clearly the full ellipse is traced out as $\phi$ varies throughout the reals because $\theta$ does so as $\phi$ does this. It is enough to just cover an interval of length $2\pi$ to get the full ellipse at least once.)$

braintic · Dec 20, 2016

Alternatively, since the Cartesian equation is given, just sub in the parametric equations and expand.

Hyperbola Locus (2 Viewers)

wrumble12

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InteGrand

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braintic

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