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 the hyperbola not containing $A$. Suppose that the triangle $ABC$ is equilateral. If the side-length of the triangle $ABC$ is $ka$, then $k$ lies in the interval

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

Answer 1

The correct option is B $(2,4]$

Given:
The equation of hyperbola,
$x^2–y^2=a^2,a>0$ and $A=(-a,0)$ and $ABC$ is a equilateral triangle.
$B=(a\sec \theta ,a\tan \theta )$ and $C=(a\sec \theta ,-a\tan \theta )$


From the figure,
$A{B}^2=B{C}^2\Rightarrow a^2(\sec \theta +1{)}^2+a^2{\tan }^2\theta =4a^2{\tan }^2\theta \Rightarrow {\sec }^2\theta +1+2\sec \theta -3{\tan }^2\theta =0\Rightarrow {\sec }^2\theta +1+2\sec \theta -3{\sec }^2\theta +3=0\Rightarrow -2{\sec }^2\theta +2\sec \theta +4=0\Rightarrow {\sec }^2\theta -\sec \theta -2=0\Rightarrow (\sec \theta -2\left)\right(\sec \theta +1)=0\Rightarrow \sec \theta =2,\sec \theta =-1$
As $\theta$ lies in first quadrant,
$\sec \theta =2\Rightarrow \tan \theta =\sqrt{{\sec }^2\theta -1}\Rightarrow \tan \theta =\sqrt{3}$
Length of the side of triangle,
$BC=2a\tan \theta =2a\sqrt{3}=ka\Rightarrow k=2\sqrt{3}$
Hence the value of $k$ lies in the interval $(2,4]$