The area (in sq units) of the equilateral triangle formed by the tangent at $(\sqrt{3}, 0)$ to the hyperbola…
The area (in sq units) of the equilateral triangle formed by the tangent at $(\sqrt{3}, 0)$ to the hyperbola $x^2-3 y^2=3$ with the pair of asymptotes of the hyperbola is
$\sqrt{2}$
$\sqrt{3}$
$\frac{1}{\sqrt{3}}$
$2 \sqrt{3}$
Solution
Given equation of hyperbola is $x^2-3 y^2=3$ $\therefore$ Equation of tangent at point $(\sqrt{3}, 0)$ is
$\begin{aligned} & S_1=0 \\ \Rightarrow \quad & x \sqrt{3}-3 y \times 0=3\end{aligned}$
$\begin{aligned} \Rightarrow & & x \sqrt{3} & =3 \\ \Rightarrow & & x & =\sqrt{3}\end{aligned}$
The asympotes of given hyperbolas are
$x+\sqrt{3} y=0$
and
$x-\sqrt{3} y=0$
On solving Eqs. (i), (ii) and (iii), we get
$(0,0),(\sqrt{3},-1)$ and $(\sqrt{3}, 1)$
$\therefore$ Area of triangle formed by joining the above points
$\begin{aligned} & =\frac{1}{2}\left|\begin{array}{rrr}0 & 0 & 1 \\ \sqrt{3} & -1 & 0 \\ \sqrt{3} & 1 & 0\end{array}\right| \\ & =\frac{1}{2}[1(\sqrt{3}+\sqrt{3})]=\frac{1}{2} \times 2 \sqrt{3}=\sqrt{3}\end{aligned}$