Let `0 lt theta lt (pi)/(2)`. If the eccentricity of the hyperbola `(x^(2))/(cos^(2)theta)-(y^(2))/(sin^(2)theta)=1` is greater than 2, then the length of its latus rectum lies in the interval

Options

A. `(1,(3)/(2)]`
B. `(3,oo)`
C. `((3)/(2),2]`
D. `(2, 3]`

Explanation

For the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`,
`e=sqrt(1+(b^(2))/(a^(2)))`
` therefore ` For the given hyperbola,
`e=sqrt(1+(sin^(2) theta)/(cos^(2)theta)) gt 2`
` " " ( because a^(2)=cos^(2) theta and b^(2)=sin^(2)theta)`
`rArr 1+tan^(2)theta gt 4`
`rArr tan^(2) theta gt 3`
`rArr tan theta in (-oo, -sqrt(3)) cup (sqrt(3), oo)`
`[x^(2) gt 3 rArr |x| gt sqrt(3) rArr x in (-oo, -sqrt(3)) cup (sqrt(3) ,oo)]`
But `theta in (0,(pi)/(2)) rArr tan theta in (sqrt(3), oo)`
`rArr theta in ((pi)/(3),(pi)/(2))`
Now, length of latusrectum
`=(2b^(2))/(a)=2(sin^(2)theta)/(cos theta)=2 sin theta tan theta `
Since, both `sin theta and tan theta ` are increasing functions in `((pi)/(3),(pi)/(2))`
` therefore ` least value of latusrectum is
`=2"sin"(pi)/(3)*"tan"(pi)/(3)=2*(sqrt(3))/(2)*sqrt(3)=3 " " ("at " theta=(pi)/(3))`
and greatest value of latusrectum is `lt oo`
Hence, latusrectum length `in(3,oo)`