Question

One focus of a hyperbola is $(3,0)$ and its corresponding directrix is $4x-3y-3=0$. If its $e=\frac{5}{4}$ then one vertex is :

• A. $\left(\frac{3}{5},\frac{11}{5}\right)$
• B. $\left(\frac{11}{5},\frac{3}{5}\right)$
• C. $(\frac{7}{5},\frac{4}{5})$
• D. $(\frac{4}{5},\frac{7}{5})$

Answer 1

Answer: (B)

$\left(\frac{11}{5},\frac{3}{5}\right)$

Distance from focus to directrix is a $(-\frac{1}{e}+e)$
$a(\frac{5}{4}-\frac{4}{5})=\frac{4\left(3\right)-3}{\mid 5\mid }$
$a\left(\frac{9}{20}\right)=\frac{9}{5}$
$\Rightarrow a=4$
Given focus $\Rightarrow$ slope of axis is $(-\frac{3}{4})$ travelling along axis parametric equation of point are $(3+r(\frac{-4}{5}\left)\right);0+r\left(\frac{3}{5}\right)$
$r=ae-a=a\left(\frac{1}{4}\right)=1$
Vertex $=(3-\frac{4}{5},\frac{3}{5})=(\frac{11}{5},\frac{3}{5})$