The equation of the tangent to the hyperbola | vedclass

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(C) The given hyperbola is $2x^2 - 3y^2 = 6$,which can be written as $\frac{x^2}{3} - \frac{y^2}{2} = 1$. Here,$a^2 = 3$ and $b^2 = 2$. The slope of t

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$y = 3x + 5$ and $y = 3x - 5$

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(C) The given hyperbola is $2x^2 - 3y^2 = 6$,which can be written as $\frac{x^2}{3} - \frac{y^2}{2} = 1$. Here,$a^2 = 3$ and $b^2 = 2$. The slope of the line $y = 3x + 4$ is $m = 3$. The equation of a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$. Substituting the values,we get $y = 3x \pm \sqrt{3(3^2) - 2}$. $y = 3x \pm \sqrt{27 - 2}$. $y = 3x \pm \sqrt{25}$. $y = 3x \pm 5$. Thus,the equations of the tangents are $y = 3x + 5$ and $y = 3x - 5$.

The equation of the tangent to the hyperbola $2x^2 - 3y^2 = 6$ which is parallel to the line $y = 3x + 4$ is:

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