Question

The equation of hyperbola whose coordinates of the foci are $(±8, 0)$ 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, $(\pm 8,0)=(\pm a e, 0)$
$\Rightarrow a e=8\,\,\,...(ii)$
and length of latusrectum $=\frac{2 b^{2}}{a}$
$\therefore 24=\frac{2 b^{2}}{a}$
$\Rightarrow b^{2}=12 a\,\,\,...(iii)$
$\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}+12 a =64 $
$ \Rightarrow a^{2}+12 a-64 =0 $
$ \Rightarrow a^{2}+16 a-4 a-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 $
$\Rightarrow 3 x^{2}-y^{2} =48$