Tangents are drawn to the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$, parallel to the straight line $2x - y = 1$. The points of contact of the tangents on the hyperbola are

Tangents are drawn to the hyperbola x²/9 - y²/4 = 1, parallel to the straight line 2x - y = 1. The points of contact of the tangents on the hyperbola are $\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1$ $2x - y = 1 \Rightarrow y = 2x - 1,\ \text{slope } m = 2$ $\text{To find: points of contact of tangents of slope } 2$ $a^2 = 9,\ b^2 = 4;\ \text{asymptotes } y = \pm\dfrac{b}{a}x = \pm\dfrac{2}{3}x$ $1.\ c^2 = a^2 m^2 - b^2$ $2.\ m = 2 \Rightarrow c = \pm 4\sqrt{2}$ $3.\ \text{contact} = \left(-\dfrac{a^2 m}{c},\ -\dfrac{b^2}{c}\right)$ $y = mx + c \text{ tangent} \iff c^2 = a^2 m^2 - b^2$ $m=2,\ a^2=9,\ b^2=4:\quad c^2 = 36 - 4 = 32$ $c = \pm 4\sqrt{2},\quad y = 2x \pm 4\sqrt{2}$ $\dfrac{xx_1}{a^2} - \dfrac{yy_1}{b^2} = 1 \Rightarrow m = \dfrac{b^2 x_1}{a^2 y_1},\ c = -\dfrac{b^2}{y_1}$ $(x_1, y_1) = \left(-\dfrac{a^2 m}{c},\ -\dfrac{b^2}{c}\right)$ $c = 4\sqrt{2}:\ x_1 = -\dfrac{18}{4\sqrt{2}} = -\dfrac{9}{2\sqrt{2}},\ y_1 = -\dfrac{4}{4\sqrt{2}} = -\dfrac{1}{\sqrt{2}}$ $c = -4\sqrt{2}:\ \left(\dfrac{9}{2\sqrt{2}},\ \dfrac{1}{\sqrt{2}}\right)$ $m = \dfrac{b^2 x_1}{a^2 y_1} = \dfrac{4}{9}\cdot\dfrac{x_1}{y_1} = \dfrac{4}{9}\cdot\dfrac{9}{2} = 2\ \checkmark$ $\boxed{(1)\ \left(\dfrac{9}{2\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right) \ \text{and}\ (2)\ \left(-\dfrac{9}{2\sqrt{2}}, -\dfrac{1}{\sqrt{2}}\right)}$