Question

The equation of the hyperbola $4x^2-32x-y^2-4y+24=0$ in its standard form is

• A. $\frac{(x-4{)}^2}{16}$- \dfrac{(y+2)^2}{36}=1$
• B. $\frac{(x-4{)}^2}{9}$- \dfrac{(y-2)^2}{36}=1$
• C. $\frac{(x-4{)}^2}{9}$- \dfrac{(y+2)^2}{36}=1$
• D. $\frac{(x-4{)}^2}{9}$- \dfrac{(y+2)^2}{16}=1$

Answer 1

The correct option is C $\frac{(x-4{)}^2}{9}$- \dfrac{(y+2)^2}{36}=1$

$4x^2-32x-y^2-4y+24=0$
$\Rightarrow 4(x^2-8x)-(y^2+4y)+24=0$
$\Rightarrow 4(x-4{)}^2-(y+2{)}^2-36=0$
$\Rightarrow$ $\frac{(x-4{)}^2}{9}$$-\frac{(y+2{)}^2}{36}=1$