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Q. Let $16\, x^2 - 3y^2 - 32x - 12y = 44$ represent a hyperbola. Then

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Solution:

Given equation of hyperbola is
$16 x^{2}-3 y^{2}-32 x-12 y=44$
$\Rightarrow 16 x^{2}+16-32 x-3 y^{2}-12-12 y$
$=44+4$
$\Rightarrow 16(x-1)^{2}-3(y+2)^{2}=48$
$\Rightarrow \frac{(x-1)^{2}}{3}-\frac{(y+2)^{2}}{16}=1$
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 $x=\pm \frac{a}{e}$
$\Rightarrow x-1=\pm \frac{\sqrt{3} \times \sqrt{3}}{\sqrt{19}}$
$\Rightarrow x=1 \pm \frac{3}{\sqrt{19}}$