Question

On a rectangular hyperbola $x^2-y^2=a^2,a>0$, three points $A,B,C$ are taken as follows : $A=(-a,0);B$ and $C$ are placed symmetrically with respect to the $X$ -axis on the branch of th e hyperbola not containing $A$. Suppose that the $\Delta ABC$ is equilateral. If the side length of the $\Delta ABC$ is $ka$, then $k$ lies in the interval

• A. (0, 2]
• B. (2, 4]
• C. (4, 6]
• D. (6, 8]

Answer 1

Answer: (B)

We have rectangular hyperbola
$x^2-y^2=a^2$

Given ABC is an equilateral triangle
$\therefore AB=BC=AC$
$A{B}^2=B{C}^2$
$a^2(\sec \theta +1{)}^2+a^2{\tan }^2\theta =4a^2{\tan }^2\theta$
$\Rightarrow (\sec \theta +1{)}^2=3{\tan }^2\theta$
$\Rightarrow (\sec \theta +1{)}^2=3({\sec }^2\theta -1)$
$\Rightarrow (\sec \theta +1{)}^2=3(\sec \theta +1)(\sec \theta -1)$
$\Rightarrow \sec \theta +1-3\sec \theta -3$
$\Rightarrow \sec \theta =2$
$\Rightarrow \theta =60^{\circ }$
$\because$ Side $BC=2a$ $\tan \theta$
$=2a\tan 60^{\circ }=2a\sqrt{3}$
But side of triangle is $ka$.
$\therefore ka-2a\sqrt{3}$
$k=2\sqrt{3}$
Hence, $k\in (2,4]$