Question

The equations of the axes of the hyperbola $9x^2-16y^2+72x-32y-16=0$ are

• A. $y+1=0$, $x+4=0$
• B. $y+2=0$, $x+3=0$
• C. $y-1=0$, $x-4=0$
• D. $y+3=0$,$x-4=0$

Answer 1

Answer: (A)

$y+1=0$, $x+4=0$

$9x^2-16y^2+72x-32y-16=0$

Completing the squares, $9(x+4{)}^2-9(16)-16(y+1{)}^2+16-16=0$

$\Rightarrow 9(x+4{)}^2-16(y+1{)}^2=9\left(16\right)$

$\Rightarrow \frac{(x+4{)}^2}{16}-\frac{(y+1{)}^2}{9}=1$

Let $x+4=X,y+1=Y$

$\Rightarrow x=X-4,Y-1$

Here $a=4,b=3$

Vertices:$(\pm a-4,-1):(\pm 4-4,-1):(0,1)$ & $(-8,-1)$

Line passing through these vertices , by two point form of line .

$y-(-1)=\frac{-1-(-1)}{-8-0}(x-0)\Rightarrow y+1=0\Rightarrow$ is the equation of one axis.

Mid point of line segment joining vertices =Centre is $(\frac{-8}{2},\frac{-1+(-1)}{2})=(-4,-1)$

The other axis will pass through (-4,-1) and will be perpendicular to the line $y+1=0$. So, line is $x=-4$.

Axes oh Hyperbola are $y=-1,x=-4$