Question

For the hyperbola $16x^2-3y^2-32x+12y-44=0,$ the correct option(s) is/are

• A. its length of transverse axis is $2\sqrt{3}$ units
• B. its length of conjugate axis is $8$ units
• C. its centre is $(1,2)$
• D. its eccentricity is $\frac{\sqrt{19}}{3}$

Answer 1

Answer: (A), (B), (C)

its centre is $(1,2)$

We have, $16(x^2-2x)-3(y^2-4y)=44$
$\Rightarrow 16(x-1{)}^2-3(y-2{)}^2=48$
$\Rightarrow \frac{(x-1{)}^2}{3}-\frac{(y-2{)}^2}{16}=1$
This equation represents a hyperbola whose centre is at $(1,2)$
and eccentricity
$e=\sqrt{1+\frac{4^2}{(\sqrt{3}{)}^2}}=\sqrt{\frac{19}{3}}$
Here,
$a^2=3\Rightarrow a=\sqrt{3},$
$b^2=16\Rightarrow b=4$
$\therefore$ Length of transverse axis $=2a=2\sqrt{3}$ units and
Length of conjugate axis $=2b=2\left(4\right)=8$ units.