Question

The foci of the hyperbola $9x^2-18x-16y^2-64y+89=0$ are

• A. $(1,3)$ , $(1,-7)$
• B. $(1,5)$ , $(1,-9)$
• C. $(1,1)$ , $(1,-5)$
• D. $(1,5)$ , $(1,-11)$

Answer 1

The correct option is A $(1,3)$ , $(1,-7)$

Given hyperbola is $9x^2-18x-16y^2-64y+89=0$
$\Rightarrow 9(x^2-2x)-16(y^2+4y)=-89$
$\Rightarrow 9(x^2-2x+1)-16(y^2+4y+4)=-89+9-64=-144$
$\Rightarrow \frac{(y+2{)}^2}{9}-\frac{(x-1{)}^2}{16}=1$
Clearly here y-axis is the transverse axis
$\Rightarrow a^2=16,b^2=9$ and $e=\sqrt{1+a^2/b^2}=\sqrt{1+16/9}=\frac{5}{3}$
Thus foci are, $x-1=0\Rightarrow x=1$ and $y+2=\pm be=\pm 5$
or $S\equiv (1,3),(1,-7)$
Hence, option 'A' is correct