Question

The area (in square units) of the equilateral triangle formed by the tangent at $(\sqrt{3},0)$ to the hyperbola $x^2-3y^2=3$ with the pair of asymptotes of the hyperbola is:

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\frac{1}{\sqrt{3}}$
• D. $2\sqrt{3}$

Answer 1

The correct option is B $\sqrt{3}$

$x^2-3y^2=3$

$\Rightarrow \frac{x^2}{3}-\frac{y^2}{1}=1$

Equation of tangent is $\frac{x{x}_1}{3}-\frac{y{y}_1}{1}=1$

$\Rightarrow x\sqrt{3}=3$

$\Rightarrow x=\sqrt{3}$ is the tangent

Asymptotes are $y=\frac{b}{a}x\Rightarrow y=\pm \frac{x}{\sqrt{3}}$
Point of intersection are $(0,0)(\sqrt{3},1)(\sqrt{3},-1)$
Hence, Area of triangle $=\frac{1}{2}base\times height=\frac{1}{2}\times \sqrt{3}\times (1+1)=\sqrt{3}$