Question

The tangent to the hyperbola $x^2-3y^2=3$ at the point $(\sqrt{3},0)$ when associated with two asymptotes constitutes

• A. scalene triangle
• B. an equilateral triangle
• C. a triangle whose area is $\sqrt{3}$ sq. units
• D. a triangle whose area is $\frac{\sqrt{3}}{2}$ sq. units

Answer 1

Answer: (B), (C)

The correct options are
B an equilateral triangle
C a triangle whose area is $\sqrt{3}$ sq. units

Tangent to the hyperbola $\frac{x^2}{3}-y^2=1$ at the point $(\sqrt{3},0),$ is given by
$\frac{\sqrt{3}x}{3}-\frac{0\times y}{1}=1$
$\Rightarrow x=\sqrt{3}\dots \left(i\right)$
Equation of asymptotes of the given hyperbola, is given by
$y=\pm \frac{b}{a}x$
$\Rightarrow y=\pm \frac{1}{\sqrt{3}}x\dots \left(ii\right)$
Solving equations $\left(i\right)$ and $\left(ii\right),$ we get vertices of the triangle as $O(0,0),A(\sqrt{3},1)$ and $B(\sqrt{3},-1).$

Here, $OA=OB=AB=2$


So, $△OAB$ is an equilateral triangle with side length of $2$ units.

Area of $△OAB$
$=\frac{\sqrt{3}}{4}\times 2^2=\sqrt{3}$ sq. units