The equation of the tangents to the conic 3x^ | vedclass

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

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$y = 3x \pm \sqrt{6}$

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(A) The given hyperbola is $3x^2 - y^2 = 3$,which can be written as $\frac{x^2}{1} - \frac{y^2}{3} = 1$. Here,$a^2 = 1$ and $b^2 = 3$. The slope of the line $x + 3y = 2$ is $m_1 = -\frac{1}{3}$. Since the tangent is perpendicular to this line,its slope $m$ must satisfy $m 	imes m_1 = -1$,so $m = 3$. The equation of a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$. Substituting $m = 3$,$a^2 = 1$,and $b^2 = 3$: $y = 3x \pm \sqrt{1(3)^2 - 3} = 3x \pm \sqrt{9 - 3} = 3x \pm \sqrt{6}$.

The equation of the tangents to the conic $3x^2 - y^2 = 3$ perpendicular to the line $x + 3y = 2$ is

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