Find the equation of Hyperbola satisfying the following conditions: Foci `(pm4,0)`, the latus rectum is of length `12`.

Explanation

Foci `(pm4,0)`, lie on x-axis.
`:." "ae=4rArra^(2)e^(2)=16` . . . (1)
and latus rectum `=(2b^(2))/(a)=12`
`rArr" "b^(2)=6a` . .(2)
`rArra^(2)(e^(2)-1)=6a`
`rArra^(2)e^(2)-a^(2)=6a`
`rArr" "16-a^(2)=6a` [From equation (1)]
`rArr" "a^(2)+6a-16=0`
`rArr" "(a+8)(a-2)=0`
`rArr" "a=-8ora=2`
But a cannot be negative.
`:.a=2rArra^(2)=4`
From equation (2)
`b^(2)=6xx2=12`
Now, equation of hyperbola
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=1rArr(x^(2))/(4)-(y^(2))/(12)=1`