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

The given equation can be written as

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

$\Rightarrow 16(x^2-2x+1-1)-3(y^2-4y+4-4)=44$

$\Rightarrow 16(x^2-2x+1)-3(y^2-4y+4)=44+16-12$

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

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

This equation represents a hyperbola with eccentricity

$e=\sqrt{1+\frac{16}{3}}=\sqrt{\frac{19}{3}}$