Q115 / 160 · WBJEE 2015 · Mathematics
Let $16 x^{2}-3 y^{2}-32 x-12 y=44$ represents a hyperbola. Then,
- Correct answerAlength of the transverse axis is $2 \sqrt{3}$
- Correct answerBlength of each latusrecturn is $32 / \sqrt{3}$
- Correct answerCeccentricity is $\sqrt{19 / 3}$
- Dequation of a directrix is $x=\frac{\sqrt{19}}{3}$
Solution
Given equation of hyperbola is $16 x^{2}-3 y^{2}-32 x-12 y=44$
$\Rightarrow 16 x^{2}+16-32 x-3 y^{2}-12-12 y$
$=44+4$
$\Rightarrow 16(x-1)^{2}-3(y+2)^{2}=48$
$\Rightarrow \quad \frac{(x-1)^{2}}{3}-\frac{(y+2)^{2}}{16}=1$
On comparing the equation with standard equation of hyperbola, we get $a=\sqrt{3}, b=4$
Now, length of transverse axis $=2 a=2 \sqrt{3}$
and length of latusrectum $=\frac{2 b^{2}}{a}=\frac{2 \times 16}{\sqrt{3}}=\frac{32}{\sqrt{3}}$
$\therefore$ Eccentricity $(e)=\sqrt{1+\frac{b^{2}}{a^{2}}}$
$=\sqrt{1+\frac{16}{3}}=\sqrt{\frac{19}{3}}$
Equation of directrix is $x=\pm \frac{a}{e}$ $\Rightarrow x-1=\pm \frac{\sqrt{3} \times \sqrt{3}}{\sqrt{19}}$
$\Rightarrow \quad x=1 \pm \frac{3}{\sqrt{19}}$
Hence, the correct answer is options: 'A', 'B', 'C'.