a m p ; x 2 9 + y 2 25 = 1 a m p ; a = 3 , b = 5 a m p ; e = 1 − 9 25 = 4 5 ∴ foci = ( 0 , ± b e ) = ( 0 , ± 4 ) a m p ; ∴ e H = 4 5 × 15 8 = 3 2 \begin{aligned}& \frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{25}=1 \\& \mathrm{a}=3, \mathrm{~b}=5 \\& \mathrm{e}=\sqrt{1-\frac{9}{25}}=\frac{4}{5} \therefore \text { foci }=(0, \pm \mathrm{be})=(0, \pm 4) \\& \quad \therefore \mathrm{e}_{\mathrm{H}}=\frac{4}{5} \times \frac{15}{8}=\frac{3}{2}\end{aligned} am p ; 9 x 2 + 25 y 2 = 1 am p ; a = 3 , b = 5 am p ; e = 1 − 25 9 = 5 4 ∴ foci = ( 0 , ± be ) = ( 0 , ± 4 ) am p ; ∴ e H = 5 4 × 8 15 = 2 3
Let equation hyperbola
a m p ; x 2 A 2 − y 2 B 2 = − 1 a m p ; ∴ B ⋅ e H = 4 ∴ B = 8 3 a m p ; ∴ A 2 = B 2 ( e H 2 − 1 ) = 64 9 ( 9 4 − 1 ) ∴ A 2 = 80 9 a m p ; ∴ x 2 80 9 − y 2 64 9 = − 1 a m p ; Directrix : y = ± B e H = ± 16 9 a m p ; P S = e ⋅ P M = 3 2 ∣ 14 3 ⋅ 2 5 − 16 9 ∣ a m p ; = 7 2 5 − 8 3 \begin{aligned}& \frac{\mathrm{x}^2}{\mathrm{~A}^2}-\frac{\mathrm{y}^2}{\mathrm{~B}^2}=-1 \\& \therefore \mathrm{B} \cdot \mathrm{e}_{\mathrm{H}}=4 \quad \therefore \mathrm{B}=\frac{8}{3} \\& \therefore \mathrm{A}^2=\mathrm{B}^2\left(\mathrm{e}_{\mathrm{H}}^2-1\right)=\frac{64}{9}\left(\frac{9}{4}-1\right) \therefore \mathrm{A}^2=\frac{80}{9} \\& \therefore \frac{\mathrm{x}^2}{\frac{80}{9}}-\frac{\mathrm{y}^2}{\frac{64}{9}}=-1 \\& \text { Directrix }: \mathrm{y}= \pm \frac{\mathrm{B}}{\mathrm{e}_{\mathrm{H}}}= \pm \frac{16}{9} \\& \mathrm{PS}=\mathrm{e} \cdot \mathrm{PM}=\frac{3}{2}\left|\frac{14}{3} \cdot \sqrt{\frac{2}{5}}-\frac{16}{9}\right| \\& =7 \sqrt{\frac{2}{5}}-\frac{8}{3}\end{aligned} am p ; A 2 x 2 − B 2 y 2 = − 1 am p ; ∴ B ⋅ e H = 4 ∴ B = 3 8 am p ; ∴ A 2 = B 2 ( e H 2 − 1 ) = 9 64 ( 4 9 − 1 ) ∴ A 2 = 9 80 am p ; ∴ 9 80 x 2 − 9 64 y 2 = − 1 am p ; Directrix : y = ± e H B = ± 9 16 am p ; PS = e ⋅ PM = 2 3 3 14 ⋅ 5 2 − 9 16 am p ; = 7 5 2 − 3 8