Question

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

• A. 2
• B. 8
• C. 10
• D. 6

Answer 1

The correct option is C

10

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

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

$\Rightarrow$ $9(x+1{)}^2-16(y-1{)}^2=9-16+151$ = 1414

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

a = 4 ,b=3

e = $\sqrt{1+\frac{9}{16}}=\frac{5}{4}$

$5S^1=2ae=\mathrm{2.4.}\frac{5}{4}$ = 10