Question

The foci of the hyperbola $9x^2-16y^2-18x+32y-151=0$ are

• A. $(1\pm 5,1)$
• B. $(1\pm 5,-1)$
• C. $(-1\pm 5,-1)$
• D. $(1\pm 5,+2)$

Answer 1

The correct option is A $(1\pm 5,1)$

$9x^2-16y^2-18x+32y-151=0$
$\Rightarrow 9(x^2-2x)-16(y^2-2y)=151$
$\Rightarrow 9(x^2-2x+1)-16(y^2-2y+1)=151+9-16=144$
$\Rightarrow \frac{(x-1{)}^2}{16}-\frac{(y-1{)}^2}{9}=1$
$\Rightarrow a^2=16,b^2=9,e=\sqrt{1+\frac{b^2}{a^2}}=\frac{5}{4}$
Therefore foci are, $(x-1)=\pm ae=\pm 5,y-1=0$
$\Rightarrow S\equiv (1\pm 5,1)$
Hence, option 'A' is correct