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
$\therefore AB=BC=AC$
$AB^{2}=BC^{2}$
$a^{2}(\sec\theta+1)^{2}+a^{2}\tan^{2}\theta=4a^{2} \,\tan^{2}\theta$
$\Rightarrow (\sec\theta+1)^{2}=3\tan^{2}\theta$
$\Rightarrow (\sec\theta+1)^{2}=3(\sec^{2} \theta-1)$
$\Rightarrow (\sec\theta+1)^{2}=3(\sec\theta+1)(\sec\theta-1)$
$\Rightarrow \sec\theta+1-3\sec\theta-3$
$\Rightarrow \sec\,\theta =2$
$\Rightarrow \theta=60^{\circ}$
$\because$ 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, $k \in (2,4]$