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 $\frac{x^2}{3} - y^2 = 1$.The equation of the tangent at point $(\

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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 $\frac{x^2}{3} - y^2 = 1$.The equation of the tangent at point $(\sqrt{3}, 0)$ is given by $x x_1 - 3 y y_1 = 3$.Substituting $(\sqrt{3}, 0)$,we get $x(\sqrt{3}) - 3y(0) = 3$,which simplifies to $x = \sqrt{3}$.The asymptotes of the hyperbola $\frac{x^2}{3} - y^2 = 1$ are given by $\frac{x^2}{3} - y^2 = 0$,which implies $x^2 = 3y^2$,or $x = \pm \sqrt{3}y$.Thus,the asymptotes are $x - \sqrt{3}y = 0$ 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:$1$. Intersection of $x = \sqrt{3}$ and $x - \sqrt{3}y = 0$: $\sqrt{3} - \sqrt{3}y = 0 \Rightarrow y = 1$. Point is $(\sqrt{3}, 1)$.$2$. Intersection of $x = \sqrt{3}$ and $x + \sqrt{3}y = 0$: $\sqrt{3} + \sqrt{3}y = 0 \Rightarrow y = -1$. Point is $(\sqrt{3}, -1)$.$3$. Intersection of $x - \sqrt{3}y = 0$ and $x + \sqrt{3}y = 0$: Point is $(0, 0)$.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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