The area (in sq units) of the triangle formed | vedclass

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(B) The given equation of the hyperbola is $x^2 - 3y^2 = 3$,which can be written as $\frac{x^2}{3} - \frac{y^2}{1} = 1$.The equation of the tangent at

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$\sqrt{3}$

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(B) The given equation of the hyperbola is $x^2 - 3y^2 = 3$,which can be written as $\frac{x^2}{3} - \frac{y^2}{1} = 1$.The equation of the tangent at the point $(\sqrt{3}, 0)$ is given by $xx_1/a^2 - yy_1/b^2 = 1$.Substituting the values,we get $x(\sqrt{3})/3 - y(0)/1 = 1$,which simplifies to $x = \sqrt{3}$.The asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by $y = \pm \frac{b}{a}x$.Here $a^2 = 3$ and $b^2 = 1$,so $a = \sqrt{3}$ and $b = 1$.The asymptotes are $y = \frac{1}{\sqrt{3}}x$ and $y = -\frac{1}{\sqrt{3}}x$,or $x - \sqrt{3}y = 0$ and $x + \sqrt{3}y = 0$.The vertices of the triangle are the intersection points of the tangent $x = \sqrt{3}$ with the asymptotes and the intersection of the two asymptotes $(0, 0)$.Intersection $1$: $x = \sqrt{3}$ and $x - \sqrt{3}y = 0$ $\Rightarrow \sqrt{3} - \sqrt{3}y = 0$ $\Rightarrow y = 1$. Point is $(\sqrt{3}, 1)$.Intersection $2$: $x = \sqrt{3}$ and $x + \sqrt{3}y = 0$ $\Rightarrow \sqrt{3} + \sqrt{3}y = 0$ $\Rightarrow y = -1$. Point is $(\sqrt{3}, -1)$.Intersection $3$: $(0, 0)$.The area of the triangle with vertices $(0, 0)$,$(\sqrt{3}, 1)$,and $(\sqrt{3}, -1)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area $= \frac{1}{2} |0(1 - (-1)) + \sqrt{3}(-1 - 0) + \sqrt{3}(0 - 1)| = \frac{1}{2} |0 - \sqrt{3} - \sqrt{3}| = \frac{1}{2} |-2\sqrt{3}| = \sqrt{3}$ sq units.

The area (in sq units) of the 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

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