Question

The hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passes through the point of intersection of the lines $7x+13y–87=0$ and $5x–8y+7=0$ and the latus rectum is $32\frac{\sqrt{2}}{5}$, then

• A. $b^2=16$
• B. $a^2=50$
• C. $a^2=\frac{25}{2}$
• D. $b^2=25$

Answer 1

Answer: (A), (C)

$a^2=\frac{25}{2}$

The point of intersection of $7x+13y-87=0$ and $5x-8y+7=0$ is $(5,4)$
Also $\frac{2b^2}{a}=32\frac{\sqrt{2}}{5}$
$b^2=\frac{16\sqrt{2}}{5}{a}^2(a>0,b>0)$
So, $\frac{x^2}{a^2}-\frac{y^2.5}{16\sqrt{2}a}=1$
It will pass through $(5,4)$
$\frac{25}{a^2}-\frac{165}{16\sqrt{2}a}=1$
$\Rightarrow \frac{5}{a^2}-\frac{1}{\sqrt{2}a}=\frac{1}{5}\Rightarrow \frac{5}{a^2}-\frac{1}{\sqrt{2}a}-\frac{1}{5}=0$
$\frac{1}{a}=\frac{\frac{1}{2}\pm \sqrt{\frac{1}{2}+4}}{2.5}=\frac{1\pm 3}{10\sqrt{2}}$
Take positive sign only
$a=\frac{10\sqrt{2}}{4}$
$a^2=\frac{25}{2}$
and $b^2=\frac{16\sqrt{2}}{5}\times \frac{5}{\sqrt{2}}=16$
$b^2=16$