Find the equation of the hyperbola, the length of whose latusrectum is 8 and eccentricity is `3//sqrt(5)dot`

Options

A. `5x^(2)-4y^(2)=100`
B. `4x^(2)-5y^(2)=100`
C. `-4x^(2)+5y^(2)=100`
D. `-5x^(2)+4y^(2)=100`

Answer

B

Explanation

Let the equation of the hyperbola be
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` ……..`(i)`
We have,
Length of the latusrectum `=8`
`implies(2b^(2))/(a)=8`
`impliesb^(2)=4a`
`impliesa^(2)(e^(2)-1)=4a`
`impliesa(e^(2)-1)=4impliesa((9)/(5)-1)=4impliesa=5`
Putting `a=5 "in" b^(2)=4a`, we get `b^(2)=20`.
Hence, the equation of the required hyperbola is `(x^(2))/(25)-(y^(2))/(20)=1`