On a rectangular hy perbola x^2-y^2= a ^2, a >0, three points A, … - Vidyadip

MCQ

On a rectangular hy perbola $x^2-y^2= a ^2, a >0$, 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 $\triangle A B C$ is equilateral. If the side length of the $\triangle A B C$ is $k a$, then $k$ lies in the interval

A
$(0,2]$
✓
$(2,4]$
C
$(4,6]$
D
$(6,8]$

✓

Answer

Correct option: B.

$(2,4]$

b
(b)

We have rectangular hyperbola

$x^2-y^2=a^2$

Given $A B C$ is an equilateral triangle.

$A B =B C=A C$

$A B^2 =B C^2$

$a^2(\sec \theta+1)^2+a^2 \tan ^2 \theta=4 a^2 \tan ^2 \theta$

$\begin{aligned}(\sec \theta+1)^2 &=3 \tan ^2 \theta \$\sec \theta+1)^2 &=3\left(\sec ^2 \theta-1\right) \$\sec \theta+1)^2 &=3(\sec \theta+1)(\sec \theta-1) \end{aligned}$

$\sec \theta+1=3 \sec \theta-3$

$\sec \theta=2$

$\theta=60^{\circ}$

$\because$ Side $B C=2 a \tan \theta$

$=2 a \tan 6 \theta^{\circ}=2 a \sqrt{3}$

But side of triangle is $k a$.

$k a =2 a \sqrt{3}$

$k =2 \sqrt{3}$

Hence, $k \in(2,4]$.

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