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 $\sqrt{\frac{19}{3}}$

Answer 1

The correct option is D Whose eccentricity is $\sqrt{\frac{19}{3}}$

Given equation is

$16(x^2-2x)-3(y^2-4y)=44$

$\Rightarrow 16(x-1{)}^2-3(y-2{)}^2=44+16-12$

$\Rightarrow 16(x-1{)}^2-3(y-2{)}^2=48$

$\Rightarrow \frac{(x-1{)}^2}{3}-\frac{(y-2{)}^2}{16}=1$

Here, centre $(1,-2)$, length of transverse axis

$=2a=2\times \sqrt{3}=2\sqrt{3}$

Length of conjugate axis $=2b=2\times 4=8$

and eccentricity, $e=\sqrt{1+\frac{b^2}{a^2}}$

$=\sqrt{1+{\left(\frac{4}{\sqrt{3}}\right)}^2}=\sqrt{1+\frac{16}{3}}$

$=\sqrt{\frac{19}{3}}$.