교점 거리 차의 극한

$\displaystyle \frac{2 t}{x} = - \frac{1}{t} x + 3$에서 $2 t ^{2} = - x ^{2} + 3 t x$ $x ^{2} - 3 t x + 2 t ^{2} = 0$, $\left( x - t \right) \left( x - 2 t \right) = 0$ 그러므로 두 점 ${\mathrm{A}}$, ${\mathrm{B}}$의 좌표는 ${\mathrm{A}} \mathit{\left( t , 2 \right)}$, ${\mathrm{B}} \mathit{\left( 2 t , 1 \right)}$ $\displaystyle \overline{{\mathrm{OA}}} = \sqrt{t ^{2} + 4}$, $\displaystyle \overline{{\mathrm{OB}}} = \sqrt{4 t ^{2} + 1}$ $\displaystyle \lim\limits _{t \rightarrow 1 +} \frac{\overline{{\mathrm{OB}}} - \overline{{\mathrm{OA}}}}{t - 1}$ $\displaystyle = \lim\limits _{t \rightarrow 1 +} \frac{\sqrt{4 t ^{2} + 1} - \sqrt{t ^{2} + 4}}{t - 1}$ $\displaystyle = \lim\limits _{t \rightarrow 1 +} \frac{\left( \sqrt{4 t ^{2} + 1} - \sqrt{t ^{2} + 4} \right) \left( \sqrt{4 t ^{2} + 1} + \sqrt{t ^{2} + 4} \right)}{\left( t - 1 \right) \left( \sqrt{4 t ^{2} + 1} + \sqrt{t ^{2} + 4} \right)}$ $\displaystyle = \lim\limits _{t \rightarrow 1 +} \frac{3 \left( t ^{2} - 1 \right)}{\left( t - 1 \right) \left( \sqrt{4 t ^{2} + 1} + \sqrt{t ^{2} + 4} \right)}$ $\displaystyle = \lim\limits _{t \rightarrow 1 +} \frac{3 \left( t + 1 \right)}{\sqrt{4 t ^{2} + 1} + \sqrt{t ^{2} + 4}} = \frac{3}{5} \sqrt{5}$ 따라서 $\displaystyle 30 \times k ^{2} = 30 \times \left( \frac{3}{5} \sqrt{5} \right) ^{2} = 54$