Q125 / 200 · WBJEE 2014 · Mathematics
The equation of hyperbola whose coordinates of the foci are (±8,0) and the length of latusrectum is 24 units, is
- Correct answerA$3 x^{2}-y^{2}=48$
- B$4 x^{2}-y^{2}=48$
- C$x^{2}-3 y^{2}=48$
- D$x^{2}-4 y^{2}=48$
Solution
Let equation of hyperbola be
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
$$
Given, foci, $(\pm 8,0)=(\pm a e, 0)$
$\Rightarrow \quad a e=8$
and length of latusrectum $=\frac{2 b^{2}}{a}$
$\therefore$
$24=\frac{2 b^{2}}{a}$
$\Rightarrow \quad b^{2}=12 a$
$\therefore$ From Eq. (ii).
$$
a^{2} e^{2}=64
$$
$\Rightarrow$
$a^{2}\left(\frac{a^{2}+b^{2}}{a^{2}}\right)=64$
$\Rightarrow \quad a^{2}+b^{2}=64$
$\Rightarrow \quad a^{2}+12 a=64$
$\Rightarrow \quad a^{2}+12 a-64=0$
$\Rightarrow \quad a^{2}+16 a-4 a-64=0$
$\Rightarrow \quad a(a+16)-4(a+16)=0$
$\Rightarrow \quad(a+16)(a-4)=0$
$\Rightarrow \quad a=4$
[$\because$ 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). $\frac{x^{2}}{16}-\frac{y^{2}}{48}=1$
$\Rightarrow \quad 3 x^{2}-y^{2}=48$
Hence, the correct answer is option: 'A'.