2006_Question2

(i) If y = 1 then the equation is:

x^2 + x + 1 = 1

giving:

x^2 + x = x(x + 1) = 0

So either x = 0 or x = -1.

(ii)

View the equation as a quadratic in x. If there are two real solutions to x, we must have a positive discriminant ie.

y^2 - 4(y^2 - 1) > 0

Giving -3y^2 + 4 > 0, or 3y^2 < 4, so:

\displaystyle -\frac {2 \sqrt 3} 3 < y < \frac {2 \sqrt 3} 3

(iii)

Again view the equation as a quadratic in x. Note that there are real solutions for x to (1) if:

\displaystyle -\frac {2 \sqrt 3} 3 \le y \le \frac {2 \sqrt 3} 3

(swap < for \le in part (ii) basically)

So the largest possible value of y is \dfrac {2 \sqrt 3} 3. We then have:

\displaystyle x^2 + \frac {2 \sqrt 3} 3 x + \frac 4 3 = 1

So:

\displaystyle x^2 + \frac {2 \sqrt 3} 3 x + \frac 1 3 = 0

note that this is a perfect square and:

\displaystyle \left(x + \frac {\sqrt 3} 3\right)^2 = 0

so:

\displaystyle x = -\frac {\sqrt 3} 3.

(iv) We have in this case:

\begin{align*}x^2 + xy + y^2 & = \left(\frac 1 {\sqrt 3} \cos \theta + \sin \theta\right)^2 + \left(\frac 1 {\sqrt 3} \cos \theta + \sin \theta\right) \left(\frac 1 {\sqrt 3} \cos \theta - \sin \theta\right) \\ &+ \left(\frac 1 {\sqrt 3} \cos \theta - \sin \theta\right)^2 \\ & = \frac 1 3 \cos^2 \theta + \frac 2 {\sqrt 3} \sin \theta \cos \theta + \sin^2 \theta + \frac 1 3 \cos^2 \theta - \sin^2 \theta + \frac 1 3 \cos^2 \theta \\ &- \frac 2 {\sqrt 3} \sin \theta \cos \theta + \sin^2 \theta \\ & = \frac 3 3 \cos^2 \theta + \sin^2 \theta \\ & = 1\end{align*}

(in fact, varying \theta, this traces out the whole curve x^2 + xy + y^2 = 1, you can see this by considering the ranges of x, y)