For conic section of the form:
![((x^2)/(a^2))-((y^2)/(b^2))=1](https://img.qammunity.org/2023/formulas/mathematics/high-school/l1mlwh7159o7ode6gyxjgegvfjx2jbjuax.png)
The Ends of the Lactus Rectum is given as:
![L=(ae,(b^2)/(a)),L=(ae,(-b^2)/(a))](https://img.qammunity.org/2023/formulas/mathematics/high-school/wgoltieqx1nyhd3x802mn3uy35g1xv4ip1.png)
The e in the equation above is the Eccentricity of the Hyperbola.
This can be obtained by the formula:
![e=\frac{\sqrt[]{a^2+b^2}}{a}](https://img.qammunity.org/2023/formulas/mathematics/high-school/w11xtyyzluuqx2mv2ktmjksqfxmfy8v2d1.png)
Thus, comparing the standard form of the conic with the given equation, we have:
![\begin{gathered} ((y+8)^2)/(16)-((x-3)^2)/(9)=1 \\ \text{This can be further expressed in the form:} \\ ((y+8)^2)/(4^2)-((x-3)^2)/(3^2)=1 \\ By\text{ comparing this with:} \\ (x^2)/(a^2)-(y^2)/(b^2)=1 \\ We\text{ can deduce that:} \\ a=4;b=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/zxs9g5w9ao7h4oq8qmtaq04po4sj1ugrkj.png)
Then, we need to obtain the value of the Eccentiricity, e.
![\begin{gathered} e=\frac{\sqrt[]{a^2+b^2}}{a} \\ e=\frac{\sqrt[]{4^2+3^2}}{4} \\ e=\frac{\sqrt[]{16+9}}{4} \\ e=\frac{\sqrt[]{25}}{4}=(5)/(4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/zlfl1pmfvndg3dmedqo17d5987d93btxip.png)
Hence, the coordinate of the ends of the each lactus rectum is:
![\begin{gathered} L=(ae,(b^2)/(a)),L=(ae,(-b^2)/(a)_{}) \\ L=(4*(5)/(4),(3^2)/(4)),L=(4*(5)/(4),(-3^2)/(4)) \\ L=(5,(9)/(4)),L=(5,(9)/(4)) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/zie8j1s1skiu3l55mehcuypz1iqmfoquj1.png)