포물선과 정사각형

$f \left( x \right) = k \left( x - 2 \right) \left( x - a \right) \left( k > 0 \right)$이라 하면 $\displaystyle f \left( x \right) = k \left( x - \frac{a + 2}{2} \right) ^{2} - \frac{k \left( a - 2 \right) ^{2}}{4}$ $\displaystyle \mathrm{P} \mathit{\left( \frac{a + 2}{2} , - \frac{k \left( a - 2 \right) ^{2}}{4} \right)}$, $\mathrm{C} \mathit{\left( 0 , 2 ak \right)}$ 사각형 $\mathrm{APRQ}$가 정사각형이므로 두 직선 $\mathrm{AP}$, ${\mathrm{BC}}$가 서로 평행하다. $\displaystyle \frac{- \frac{k \left( a - 2 \right) ^{2}}{4}}{\frac{a + 2}{2} - 2} = \frac{- 2 ak}{a}$ $\displaystyle \frac{- k \left( a - 2 \right)}{2} = - 2 k$, $a = 6$ $\mathrm{P} \mathit{\left( 4 , - 4 k \right)}$, $\mathrm{C} \mathit{\left( 0 , 12 k \right)}$ 직선 $\mathrm{BC}$의 방정식은 $2 k x + y - 12 k = 0$ 사각형 $\mathrm{APRQ}$가 정사각형이므로 $\displaystyle {\overline{\mathrm{AP}}} = {\overline{\mathrm{AQ}}}$ $\displaystyle \sqrt{2 ^{2} + \left( - 4 k \right) ^{2}} = \frac{\left| 4 k - 12 k \right|}{\sqrt{\left( 2 k \right) ^{2} + 1 ^{2}}}$ $\displaystyle \sqrt{4 \left( 4 k ^{2} + 1 \right)} = \frac{8 k}{\sqrt{4 k ^{2} + 1}}$ $4 k ^{2} + 1 = 4 k$ $\left( 2 k - 1 \right) ^{2} = 0$, $\displaystyle k = \frac{1}{2}$ $\displaystyle f \left( x \right) = \frac{1}{2} \left( x - 2 \right) \left( x - 6 \right)$ 따라서 $f \left( 12 \right) = 30$ [참고]