Question

find the equ. of hyperbole lactus rectum is 8 and e is 3/root 5 ​

Answer 1

Step-by-step explanation:

We have,

Latus rectum = 8

Eccentricity,

$e = \frac{3}{ \sqrt{5} }$

$\implies {e}^{2} = \frac{9}{5}$

$\implies1 + \frac{ {b}^{2} }{ {a}^{2} } = \frac{9}{5}$

$\implies \frac{ {b}^{2} }{ {a}^{2} } = \frac{4}{5}$

$\implies \frac{b}{a} = \frac{2}{ \sqrt{5} }$

Also,

$2\frac{ {b}^{2} }{a} = 8$

$\implies \frac{ {b}^{2} }{a} = 4$

Now,

$\frac{b}{a} \times \frac{a}{ {b}^{2} } = \frac{2}{ \sqrt{5} } \times \frac{1}{4}$

$\implies \frac{1}{b} = \frac{1}{2 \sqrt{5} }$

$\implies \: b = 2 \sqrt{5}$

$a = \frac{ \sqrt{5} }{2} \times b$

$\implies \: a = \frac{ \sqrt{5} }{2} \times 2 \sqrt{5}$

$\implies \: a = 5$

Required equation of hyperbola,

$\frac{ {x}^{2} }{25} - \frac{ {y}^{2} }{20} = 1$

$\implies4 {x}^{2} - 5 {y}^{2} = 100$