Question

The equation $16x^2-3y^2-32x+12y-44=0$ represents a hyperbola

• A. the length of whose transverse axis is $4\sqrt{3}$
• B. the length of whose conjugate axis is 4
• C. whose centre is (-1,2)
• D. whose eccentricity is $\sqrt{19/3}$

Answer 1

Answer: (D)

The given equation can be written as
$16\left(x^2-2x\right)-3\left(y^2-4y\right)-44=0$
i.e., $16{\left(x-1\right)}^2-3{\left(y-2\right)}^2=44+16-12$
is given by ${}^{\prime \prime }T=S_1^{\prime \prime }$ i.e.,
$\Rightarrow \frac{{\left(x-1\right)}^2}{3}-\frac{{\left(y-2\right)}^2}{16}=1$
which is a hyperbola whose eccentricity is given by
$b^2=a^2\left(e^2-1\right)$
$\Rightarrow 16=3\left(e^2-1\right)$
$\Rightarrow e^2-1=\frac{16}{13}$
$\Rightarrow e^2=1+\frac{16}{3}=\frac{19}{3}$
$\Rightarrow e=\sqrt{19/3}$