Question

find the coordinates of the ends of each latus rectum and equations of asymptotes.

Answer 1

For conic section of the form:

$(\frac{x^2}{a^2})-(\frac{y^2}{b^2})=1$

The Ends of the Lactus Rectum is given as:

$L=(ae,\frac{b^2}{a}),L=(ae,\frac{-b^2}{a})$

The e in the equation above is the Eccentricity of the Hyperbola.

This can be obtained by the formula:

$e=\frac{\sqrt[]{a^2+b^2}}{a}$

Thus, comparing the standard form of the conic with the given equation, we have:

$\begin{gathered} \frac{(y+8)^2}{16}-\frac{(x-3)^2}{9}=1 \\ \text{This can be further expressed in the form:} \\ \frac{(y+8)^2}{4^2}-\frac{(x-3)^2}{3^2}=1 \\ By\text{ comparing this with:} \\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ We\text{ can deduce that:} \\ a=4;b=3 \end{gathered}$

Then, we need to obtain the value of the Eccentiricity, e.

$\begin{gathered} e=\frac{\sqrt[]{a^2+b^2}}{a} \\ e=\frac{\sqrt[]{4^2+3^2}}{4} \\ e=\frac{\sqrt[]{16+9}}{4} \\ e=\frac{\sqrt[]{25}}{4}=\frac{5}{4} \end{gathered}$

Hence, the coordinate of the ends of the each lactus rectum is:

$\begin{gathered} L=(ae,\frac{b^2}{a}),L=(ae,\frac{-b^2}{a}_{}) \\ L=(4\times\frac{5}{4},\frac{3^2}{4}),L=(4\times\frac{5}{4},\frac{-3^2}{4}) \\ L=(5,\frac{9}{4}),L=(5,\frac{9}{4}) \end{gathered}$