The distance between the tangents to the hype | vedclass

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(B) The equation of the hyperbola is $\frac{x^2}{20} - \frac{y^2}{4/3} = 1$. Here $a^2 = 20$ and $b^2 = \frac{4}{3}$.Given the line $x + 3y = 7$,the s

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$\frac{4}{\sqrt{5}}$

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(B) The equation of the hyperbola is $\frac{x^2}{20} - \frac{y^2}{4/3} = 1$. Here $a^2 = 20$ and $b^2 = \frac{4}{3}$.Given the line $x + 3y = 7$,the slope $m = -\frac{1}{3}$.The equation of the tangent to the hyperbola in slope form is $y = mx \pm \sqrt{a^2m^2 - b^2}$.Substituting the values: $y = -\frac{1}{3}x \pm \sqrt{20(-\frac{1}{3})^2 - \frac{4}{3}} = -\frac{1}{3}x \pm \sqrt{\frac{20}{9} - \frac{12}{9}} = -\frac{1}{3}x \pm \sqrt{\frac{8}{9}} = -\frac{1}{3}x \pm \frac{2\sqrt{2}}{3}$.This gives the two parallel tangents: $x + 3y - 2\sqrt{2} = 0$ and $x + 3y + 2\sqrt{2} = 0$.The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.$d = \frac{|2\sqrt{2} - (-2\sqrt{2})|}{\sqrt{1^2 + 3^2}} = \frac{4\sqrt{2}}{\sqrt{10}} = \frac{4\sqrt{2}}{\sqrt{2}\sqrt{5}} = \frac{4}{\sqrt{5}}$.

The distance between the tangents to the hyperbola $\frac{x^2}{20} - \frac{3y^2}{4} = 1$ which are parallel to the line $x + 3y = 7$ is

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