Parameter

Geometry Level pending

$\begin{cases} x=\dfrac{4\alpha }{1+\alpha ^2} \\ y=\dfrac{2-2\alpha ^2}{1+\alpha ^2} \end{cases}$

If reals $x$ and $y$ satisfy the system of equations above for real parameter $\alpha$ . Find the range $x^2-xy+y^2$ lies within.

[2, 6] [4, 6] None of the others [2, 4]

1 solution

Chew-Seong Cheong
Oct 3, 2016

Let $\alpha = \tan \theta$ . Then we have:

$\begin{cases} x = \dfrac {4\alpha}{1+\alpha^2} = \dfrac {2(2\tan \theta)}{1+\tan^2 \theta} = 2 \sin 2\theta \\ y = \dfrac {2-2\alpha^2}{1+\alpha^2} = \dfrac {2(1-\tan^2 \theta)}{1+\tan^2 \theta} = 2 \cos 2\theta \end{cases} \quad$ by Weierstrass substitution or tangent half-angle substitution .

Therefore,

$\begin{aligned} x^2 - xy + y^2 & = 4\sin^2 2\theta - 4\sin 2\theta \cos 2\theta + 4\cos^2 2\theta \\ & = 4 - 2\sin 4\theta \end{aligned}$

Since $\sin 4\theta \in [-1, 1] \implies x^2 - xy + y^2 \in \boxed{[2,6]}$ .

Sahil, need not key in text in LaTex, it is difficult, does not look good and not the standard practice. The standard in Brilliant is to just key them as text. I have edited the problem for you.

Chew-Seong Cheong - 4 years, 8 months ago

Thanks sir!

Sahil Silare - 4 years, 8 months ago

There's mistake in your step. You have written $4\cos ^22\theta$ as $4\cos ^2\theta$

Sahil Silare - 4 years, 8 months ago

Thanks, I have changed it.

Chew-Seong Cheong - 4 years, 8 months ago

Parameter

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