If a directrix of a hyperbola centred at the origin and passing through the point `(4,-2 sqrt(3)) " is " 5x=4sqrt(5)` and its eccentricity is e, then

Options

A. `4e^(4)-12e^(2)-27=0`
B. `4e^(4)-24e^(2)+27=0`
C. `4e^(4)+8e^(2)-35=0`
D. `4e^(4)-24e^(2)+35=0`

Explanation

Let the equation of hyperbola is
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=1 " ...(i)" `
Since, equation of given directrix is `5x=4sqrt(5)`
so `5((a)/(e))=4sqrt(5) " "[ because " equation of directrix is "x=(a)/(e)]`
`rArr (a)/(e)=(4)/(sqrt(5)) " ...(ii)" `
and hyperbola (i) passes through point `(4-2 sqrt(3))`
so, `(16)/(a^(2))-(12)/(b^(2))=1 " ...(iii)" `
The eccentricity `e=sqrt(1+(b^(2))/(a^(2)))`
` rArr e^(2) =1+(b^(2))/(a^(2))`
`rArr a^(2)e^(2)-a^(2)=b^(2) " ...(iv)" `
From Eqs. (ii) and (iv) , we get
`(16)/(5)e^(4)-(16)/(5)e^(2)=b^(2) " ...(v)" `
From Eqs. (ii) and (iii), we get
`(16)/((16)/(5)e^(2))-(12)/(b^(2))=1 rArr (5)/(e^(2))-(12)/(b^(2))=1`
`rArr (12)/(b^(2))=(5)/(e^(2))-1 rArr (12)/(b^(2))=(5-e^(2))/(e^(2))`
`rArr b^(2)=(12e^(2))/(5-e^(2)) " ...(vi)" `
From Eqs. (v) and (vi), we get
`16 e^(4)-16e^(2)=5((12e^(2))/(5-e^(2))) rArr 16(e^(2)-1)(5-e^(2))=60`
`rArr 4(5e^(2)-e^(4)-5+e^(2))=15`
` rArr 4e^(4)-24e^(2) +35=0`