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Question

(B) The given equation of the 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 $b^2 = 3$.

Vedclass · Accepted Answer

$\frac{2}{\sqrt{5}}$

Vedclass · Answer

(B) The given equation of the 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 equation of a tangent with slope $m$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$. Given the line is $y = 2x + 4$,the slope $m = 2$. Substituting $m = 2, a^2 = 1, b^2 = 3$ into the tangent equation: $y = 2x \pm \sqrt{1(2)^2 - 3} = 2x \pm \sqrt{4 - 3} = 2x \pm 1$. The two parallel tangents are $2x - y + 1 = 0$ and $2x - y - 1 = 0$. The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. $d = \frac{|1 - (-1)|}{\sqrt{2^2 + (-1)^2}} = \frac{2}{\sqrt{4 + 1}} = \frac{2}{\sqrt{5}}$.

The distance between the tangents drawn to the hyperbola $3x^2 - y^2 = 3$,which are parallel to the line $y = 2x + 4$,is:

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