Question

The foci of the hyperbola $16x^2-9y^2-64x+18y-90=0$ are

• A. $\left(\frac{24\pm 5\sqrt{145}}{12},1\right)$
• B. $\left(\frac{21\pm 5\sqrt{145}}{12},1\right)$
• C. $\left(1,\frac{24\pm 5\sqrt{145}}{2}\right)$
• D. $\left(1,\frac{21\pm 5\sqrt{145}}{2}\right)$
• E. $\left(\frac{21\pm 5\sqrt{145}}{2},-1\right)$

Answer 1

Answer: (A)

$16x^2-9y^2-64x+18y-90=0$
$-16\left(x^2-4x\right)-9\left(y^2-2y\right)-90$
$=16(x-2{)}^2-9(y-1{)}^2=90+16\times 4-9\times 1$
$=16(x-2{)}^2-9(y-1{)}^2=145$
$=\frac{(x-2{)}^2}{\frac{145}{16}}-\frac{(y-1{)}^2}{\frac{145}{9}}=1$ ...(i)
We know that, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ...(ii)
Foci $\Rightarrow (ae,0)$
On comparing Eqs. (i) and (ii), we get
$\because e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$
$X=ae\Rightarrow x-2$
$=+\sqrt{\frac{145}{16}}\times \frac{5}{3}$
$=\pm \frac{5\sqrt{145}}{12}$
$x=2\pm \frac{5\sqrt{145}}{12}$
$\Rightarrow x=\frac{24\pm 5\sqrt{145}}{12}$
$Y=0\Rightarrow y-1=0$
$\Rightarrow y=1$
Hence, $\left(\frac{24\pm 5\sqrt{145}}{12},1\right)$