The tangent to the hyperbola x^2 - 3y^2 = 3 a | vedclass

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(C) The given hyperbola is $x^2 - 3y^2 = 3$,which can be written as $\frac{x^2}{3} - \frac{y^2}{1} = 1$.Here,$a^2 = 3$ and $b^2 = 1$,so $a = \sqrt{3}$

Vedclass · Accepted Answer

both $(A)$ and $(B)$

Vedclass · Answer

(C) The given hyperbola is $x^2 - 3y^2 = 3$,which can be written as $\frac{x^2}{3} - \frac{y^2}{1} = 1$.Here,$a^2 = 3$ and $b^2 = 1$,so $a = \sqrt{3}$ and $b = 1$.The point $(\sqrt{3}, 0)$ is the vertex of the hyperbola.The equation of the tangent at $(\sqrt{3}, 0)$ is $x = \sqrt{3}$.The asymptotes of the hyperbola are $y = \pm \frac{b}{a}x$,which are $y = \pm \frac{1}{\sqrt{3}}x$.To find the vertices of the triangle formed by the tangent $x = \sqrt{3}$ and the asymptotes $y = \frac{1}{\sqrt{3}}x$ and $y = -\frac{1}{\sqrt{3}}x$:$1$. Intersection of $x = \sqrt{3}$ and $y = \frac{1}{\sqrt{3}}x$ is $(\sqrt{3}, 1)$.$2$. Intersection of $x = \sqrt{3}$ and $y = -\frac{1}{\sqrt{3}}x$ is $(\sqrt{3}, -1)$.$3$. Intersection of the two asymptotes is $(0, 0)$.The vertices of the triangle are $(0, 0)$,$(\sqrt{3}, 1)$,and $(\sqrt{3}, -1)$.The base of the triangle along the line $x = \sqrt{3}$ has length $|1 - (-1)| = 2$.The height of the triangle from the origin to the line $x = \sqrt{3}$ is $\sqrt{3}$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 	imes \sqrt{3} = \sqrt{3}$ sq. units.The sides of the triangle are $2$,$\sqrt{(\sqrt{3})^2 + 1^2} = 2$,and $\sqrt{(\sqrt{3})^2 + (-1)^2} = 2$. Since all sides are equal,it is an equilateral triangle.Thus,both $(A)$ and $(B)$ are correct.

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

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