Chapter A5.2, Problem 23E

To determine

To find:the equation of hyperbola.

Answer to Problem 23E

The equation of hyperbolaas: ${\frac{\left(y-6\right)}{36}}^2-\frac{{\left(x-1\right)}^2}{45}=1$

Explanation of Solution

Given:

Focus $\left(1,-3\right)$

Eccentricity $e=\frac{3}{2}$

Directrix $y=2$

Calculation:

We are given with focus of hyperbola as $\left(1,-3\right)$ ,eccentricity $e=\frac{3}{2}$ and directrix as $y=2$ .

We can find the equation of hyperbola as:

  $\begin{array}{l}\Rightarrow \sqrt{{\left(x-1\right)}^2+{\left(y+3\right)}^2}=\frac{3}{2}\left|y-2\right|\\ \Rightarrow {\left(x-1\right)}^2+{\left(y+3\right)}^2=4{\left(y-2\right)}^2\\ \Rightarrow 4\left(x^2-2x+1\right)-5\left(y^2-12y+36\right)=-4+4-180\\ \Rightarrow {\frac{\left(y-6\right)}{36}}^2-\frac{{\left(x-1\right)}^2}{45}=1\end{array}$

Conclusion:

Therefore,the equation of hyperbola as: ${\frac{\left(y-6\right)}{36}}^2-\frac{{\left(x-1\right)}^2}{45}=1$