Find the equation of the hyperbola with foci `(0, pm10)` and 30 as the length of latus rectum.

Explanation

Since the foci are on the y-axis, the required equation of the hyperbola would be of the form `(y^(2))/(a^(2)) - (x^(2))/(b^(2)) =1`
Now, co-ordinates of the foci are `(0 , pm 10) rArr c = 10`
Length of the latus rectum `= (2b^(2))/(a) = 30 rArr b^(2) = 15a`
We, know that `c^(2) = a^(2) + b^(2)" " ...(i) `
`rArr (10)^(2) = a^(2) + 15a`
`rArr a^(2) + 15a - 100 = 0`
`rArr a^(2) = 20a - 5a - 100 = 0`
`rArr a(a+20) -5(a + 20) = 0`
`rArr a = 5" "...(ii)" " ` [As a cannot be negative]
Equation (i) and (ii) implies `c^(2) = 25 + b^(2)`
`rArr (10)^(2) = 25 + b^(2)`
`rArr b^(2) = 100-25 = 75`
`therefore (y^(2))/(25)-(x^(2))/(75) = 1` is the required equation of the hyperbola.