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**Prove It** $\displaystyle \displaystyle \begin{align*} \sqrt{(x + 1)^2 + y^2} - \sqrt{(x - 2)^2 + y^2} &= 1 \\ \left(\sqrt{(x + 1)^2 + y^2} - \sqrt{(x - 2)^2 + y^2}\right)^2 &= 1^2 \\ (x + 1)^2 + y^2 - 2\sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2} + (x - 2)^2 + y^2 &= 1 \\ (x + 1)^2 + (x - 2)^2 + 2y^2 - 1 &= 2\sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2} \\ x^2 + 2x + 1 + x^2 - 4x + 4 + 2y^2 - 1 &= 2\sqrt{(x+1)^2 + y^2}\sqrt{(x-2)^2 + y^2} \\ 2x^2 - 2x + 4 + 2y^2 &= 2\sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2} \\ x^2 - x + 2 + y^2 &= \sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2} \\ \left(x^2 - x + 2 + y^2\right)^2 &= \left(\sqrt{(x + 1)^2 + y^2}\sqrt{(x - 2)^2 + y^2}\right)^2 \\ x^4-2x^3+2x^2y^2+5x^2-2xy^2-4x+y^4+4y^2+4 &= \left[(x + 1)^2 + y^2\right]\left[(x - 2)^2 + y^2\right] \\ x^4-2x^3+2x^2y^2+5x^2-2xy^2-4x+y^4+4y^2+4 &= x^4-2x^3+2x^2y^2-3x^2-2xy^2+4x+y^4+5y^2+4 \\ 5x^2 - 4x + 4y^2 &= -3x^2 + 4x + 5y^2\end{align*}$

$\displaystyle \displaystyle \begin{align*} 8x^2 - 8x - y^2 &= 0 \\ 8\left(x^2 - x\right) - y^2 &= 0 \\ 8\left[x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2\right] - y^2 &= 0 \\ 8\left[\left(x - \frac{1}{2}\right)^2 - \frac{1}{4}\right] - y^2 &= 0 \\ 8\left(x - \frac{1}{2}\right)^2 - y^2 - 2 &= 0 \\ 8\left(x - \frac{1}{2}\right)^2 - y^2 &= 2 \\ 4\left(x - \frac{1}{2}\right)^2 - \frac{y^2}{2} &= 1 \\ \frac{\left(x - \frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2} - \frac{y^2}{\left(\sqrt{2}\right)^2} &= 1\end{align*}$

So this is a hyperbola, centred at $\displaystyle \displaystyle (x, y) = \left(\frac{1}{2}, 0\right)$.