Question

The equation of hyperbola whose coordinates of the foci are $(\pm 8,0)$ and the length of latus rectum is $24$ units, is

• A. $3x^2-y^2=48$
• B. $4x^2-y^2=48$
• C. $x^2-3y^2=48$
• D. $x^2-4y^2=48$

Answer 1

Answer: (A)

Let equation of hyperbola be
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1...(i)$
Given, foci, $(\pm 8,0)=(\pm ae,0)$
$\Rightarrow ae=8...(ii)$
and length of latusrectum $=\frac{2b^2}{a}$
$\therefore 24=\frac{2b^2}{a}$
$\Rightarrow b^2=12a...(iii)$
$\therefore$ From Eq. (ii),
$a^2{e}^2=64$
$\Rightarrow a^2\left(\frac{a^2+b^2}{a^2}\right)=64$
$\Rightarrow a^2+b^2=64$
$\Rightarrow a^2+12a=64$
$\Rightarrow a^2+12a-64=0$
$\Rightarrow a^2+16a-4a-64=0$
$\Rightarrow a(a+16)-4(a+16)=0$
$\Rightarrow (a+16)(a-4)=0$
$\Rightarrow a=4$
$[\because$ a cannot be negative]
On putting $a=4$ in Eq. (iii), we get
$b^2=12\times 4$
$\Rightarrow b^2=48$
$\therefore$ From Eq. (i),
$\frac{x^2}{16}-\frac{y^2}{48}=1$
$\Rightarrow 3x^2-y^2=48$