The equations of the tangents to the hyperbol | vedclass

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(D) The given hyperbola is $4x^2 - y^2 = 12$,which can be written as $\frac{x^2}{3} - \frac{y^2}{12} = 1$.Here,$a^2 = 3$ and $b^2 = 12$.The equation o

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$12$

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(D) The given hyperbola is $4x^2 - y^2 = 12$,which can be written as $\frac{x^2}{3} - \frac{y^2}{12} = 1$.Here,$a^2 = 3$ and $b^2 = 12$.The equation of a tangent with slope $m$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$.Given the slope $m = 4$,we substitute the values:$c = \pm \sqrt{3(4)^2 - 12} = \pm \sqrt{3(16) - 12} = \pm \sqrt{48 - 12} = \pm \sqrt{36} = \pm 6$.Thus,the equations are $y = 4x + 6$ and $y = 4x - 6$,so $c_1 = 6$ and $c_2 = -6$.Therefore,$|c_1 - c_2| = |6 - (-6)| = |12| = 12$.

The equations of the tangents to the hyperbola $4x^2 - y^2 = 12$ are $y = 4x + c_1$ and $y = 4x + c_2$. Then $|c_1 - c_2|$ is equal to -

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