Question

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

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