The distance between the tangents of the hype | vedclass

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(A) 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 th

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$2\sqrt{2}$

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(A) 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 $x - 2y + 5 = 0$ is $m_1 = \frac{1}{2}$.The tangents are perpendicular to this line,so their slope $m$ must satisfy $m 	imes \frac{1}{2} = -1$,which gives $m = -2$.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 $m = -2, a^2 = 3, b^2 = 2$,we get $y = -2x \pm \sqrt{3(-2)^2 - 2} = -2x \pm \sqrt{12 - 2} = -2x \pm \sqrt{10}$.So,the two tangents are $2x + y - \sqrt{10} = 0$ and $2x + y + \sqrt{10} = 0$.The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.Here,$d = \frac{|\sqrt{10} - (-\sqrt{10})|}{\sqrt{2^2 + 1^2}} = \frac{2\sqrt{10}}{\sqrt{5}} = 2\sqrt{2}$.

The distance between the tangents of the hyperbola $2x^2 - 3y^2 = 6$ which are perpendicular to the line $x - 2y + 5 = 0$ is

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