Let $x+y+1=0$ and $x-y+4=0$ be the asymptotes of a hyperbola $H$. If $(1,1)$ is a point on $H$, then the…
Let $x+y+1=0$ and $x-y+4=0$ be the asymptotes of a hyperbola $H$. If $(1,1)$ is a point on $H$, then the length of the latus rectum of $\mathrm{H}$ is
$4 \sqrt{3}$
$\sqrt{3}$
$4 \sqrt{2}$
$\sqrt{5}$
Solution
We know that, the a symptotes of hyperbola $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ is
$(y-k)= \pm \frac{b}{a}(x-h)$... (i)
and latus rectum is $\frac{2 b^2}{a}$
Rewrite the given a symptotes are
$
\begin{aligned}
& \left(y-\frac{3}{2}\right)= \pm 1\left(x+\frac{5}{2}\right) \Rightarrow \frac{b}{a}=1 \\
& \Rightarrow \frac{\left(x+\frac{5}{2}\right)^2}{a^2}-\frac{\left(y-\frac{3}{2}\right)^2}{b^2}=1
\end{aligned}
$
Since $b=a$ and hyperbola passes through $(1,1)$. Hence
$
\Rightarrow a^2=\left(1+\frac{5}{2}\right)^2-\left(1-\frac{3}{2}\right)^2=12
$
Hence $b^2=a^2=12$
Therefore latus rectum $=2\left(\frac{b^2}{a}\right)=2\left(\frac{12}{\sqrt{12}}\right)=4 \sqrt{3}$