쌍곡선과 수직 초점반지름

$\displaystyle {\overline{\mathrm{PF}}} = t$라 하자. 쌍곡선의 정의에 의하여 $\displaystyle {\overline{\mathrm{PF} \mathit{'}}} - {\overline{\mathrm{PF}}} = 2 a$ $\displaystyle {\overline{\mathrm{PF} \mathit{'}}} = t + 2 a$ $\displaystyle {\overline{\mathrm{AF}}} = {\overline{\mathrm{AF} \mathit{'}}}$, $\displaystyle {\overline{\mathrm{AR}}} = {\overline{\mathrm{AS}}}$, $\displaystyle \angle \mathrm{ARF} \mathit{=} \angle \mathrm{ASF} \mathit{'} = \frac{\pi}{2}$이므로 두 삼각형 $\mathrm{ARF}$, $\mathrm{ASF} \mathit{'}$은 서로 합동이고 $\displaystyle {\overline{\mathrm{RF}}} = {\overline{\mathrm{SF} \mathit{'}}}$ $\displaystyle {\overline{\mathrm{AR}}} = {\overline{\mathrm{AS}}} = k$라 하면 $\displaystyle {\overline{\mathrm{RF}}} = t + k$, $\displaystyle {\overline{\mathrm{SF} \mathit{'}}} = {\overline{\mathrm{PF} \mathit{'}}} - {\overline{\mathrm{PS}}} = t + 2 a - k$ $t + k = t + 2 a - k$ $k = a$ $\displaystyle {\overline{\mathrm{PQ}}} = \frac{a}{3}$이므로 $\displaystyle {\overline{\mathrm{QS}}} = \frac{2 a}{3}$ 두 삼각형 $\mathrm{QSA}$, $\mathrm{QPF}$는 서로 닮음이고 닮음비가 $2 : 1$이므로 $\displaystyle {\overline{\mathrm{PF}}} = \frac{1}{2} {\overline{\mathrm{AS}}} = \frac{a}{2}$ $\displaystyle {\overline{\mathrm{PF} \mathit{'}}} = {\overline{\mathrm{PF}}} + 2 a = \frac{5}{2} a$ $\displaystyle \angle \mathrm{F} \mathit{'} \mathrm{PF} \mathit{=} \frac{\pi}{2}$, $\displaystyle {\overline{\mathrm{PF}}} = \frac{a}{2}$, $\displaystyle {\overline{\mathrm{PF} \mathit{'}}} = \frac{5}{2} a$이므로 $\displaystyle {\overline{\mathrm{FF} \mathit{'}}} = \frac{\sqrt{26}}{2} a$ $\cdots$$\cdots$ ㉠ 두 점 $\mathrm{F}$, $\mathrm{F} \mathit{'}$은 쌍곡선 $\displaystyle \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{3} = 1$의 두 초점이므로 $\displaystyle {\overline{\mathrm{FF} \mathit{'}}} = 2 \sqrt{a ^{2} + 3}$ $\cdots$$\cdots$ ㉡ 두 식 ㉠, ㉡에 의하여 $26 a ^{2} = 16 a ^{2} + 48$ 따라서 $\displaystyle a ^{2} = \frac{24}{5}$