Question

The equation of the transverse and conjugate axis of the hyperbola $16x^2-y^2+64x+4y+44=0$ are

• A. $x=2,y+2=0$
• B. $x=2,y=2$
• C. $y=2,x+2=0$
• D. $Noneofthese$

Answer 1

The correct option is C $y=2,x+2=0$

$\begin{array}{c}16x^2-y^2+64x+4y+44=0\\ \Rightarrow 16x^2+64x-\left(y^2-4y\right)+44=0\\ \Rightarrow 16\left(x^2+4x+4-4\right)-\left(y^2-4y+4-4\right)+44=0\\ \Rightarrow 16{\left(x+2\right)}^2-64-{\left(y-2\right)}^2+4+44=0\\ \Rightarrow 16{\left(x+2\right)}^2-{\left(y-2\right)}^2-16=0\\ {\left(x+2\right)}^2-\frac{{\left(y-2\right)}^2}{16}=1\end{array}$

$\therefore$ Trans-verse axis $y=2$, and conjugate axis $x+2=0$.