Tangents are drawn to the hyperbola frac{x^2} | vedclass

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(D) The slope of the line $2x - y = 1$ is $m = 2$.The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$

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$(A, B)$

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(D) The slope of the line $2x - y = 1$ is $m = 2$.The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.Here $a^2 = 9$ and $b^2 = 4$,so the tangent equations are $y = 2x \pm \sqrt{9(2)^2 - 4} = 2x \pm \sqrt{36 - 4} = 2x \pm \sqrt{32} = 2x \pm 4\sqrt{2}$.The point of contact $(x_1, y_1)$ for a tangent $y = mx + c$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\left(\frac{a^2m}{c}, \frac{b^2}{c}\right)$.For $c = 4\sqrt{2}$,the point is $\left(\frac{9(2)}{4\sqrt{2}}, \frac{4}{4\sqrt{2}}\right) = \left(\frac{18}{4\sqrt{2}}, \frac{1}{\sqrt{2}}\right) = \left(\frac{9}{2\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$,which is point $(A)$.For $c = -4\sqrt{2}$,the point is $\left(\frac{9(2)}{-4\sqrt{2}}, \frac{4}{-4\sqrt{2}}\right) = \left(-\frac{9}{2\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$,which is point $(B)$.Thus,the points of contact are $(A)$ and $(B)$.

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