Question

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

• A. $(4,1)$ and $(-6,1)$
• B. $(-4,1)$ and $(-6,1)$
• C. $(-4,1)$ and $(6,-1)$
• D. $(4,1)$ and $(6,-1)$

Answer 1

The correct option is A $(4,1)$ and $(-6,1)$

$9x^2-16y^2+18x+32y-151=0$

$\Rightarrow 9(x^2+2x)-16(y^2-2y)-151=0$

$\Rightarrow 9(x^2+2x+1-1)-16(y^2-2y+1-1)-151=0$

$\Rightarrow 9{(x+1)}^2-9-16{(y-1)}^2+16-151=0$

$\Rightarrow 9{(x+1)}^2-16{(y-1)}^2=144$

$\Rightarrow \frac{(x+1)62}{16}-\frac{{(y-1)}^2}{9}=1$

Let $x+1=X$ and $y-1=Y$

then we have the equation of hyperbola in the standard form in the transformed coordinates -

$\Rightarrow \frac{X^2}{16}-\frac{Y^2}{9}=1$

where the length of major axis is $2a=8$

and length of minor axis is $2b=6$

so, the coordinates of the focii will be $(\pm ae,o)$ where $e$ is the eccentricity.

we have, $\frac{b^2}{a^2}=e^2-1$

$\Rightarrow \frac{9}{16}=e^2-1$

$\Rightarrow e^2=\frac{25}{16}$

$\Rightarrow e=\frac{5}{4}$

$\therefore$ Coordinates of focii are $(\pm 5,0)$

However in the $xy$ coordinate system will be $(5-1,1)$ and $(-5-1,1)$

$=(4,1)$ and $(-6,1)$

Hence, the answer is $(4,1)$ and $(-6,1).$