Find the equation of the two tangents that can be drawn from the point $\left(1, 2\right)$ to the hyperbola $2x^2 - 3y^2 = 6$.
Equation of hyperbola is $\;$ $2x^2 - 3y^2 = 6$
i.e. $\;$ $\dfrac{x^2}{6 / 2} - \dfrac{y^2}{6 / 3} = 1$
i.e. $\;$ $\dfrac{x^2}{3} - \dfrac{y^2}{2} = 1$ $\;\;\; \cdots \; (1)$
The transverse axis of the hyperbola given by equation $(1)$ is along the X axis.
$\therefore$ $\;$ Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives
$a^2 = 3$ $\;$ and $b^2 = 2$
Let the equation of the tangent be $\;$ $y = mx + \sqrt{a^2 m^2 - b^2}$ $\;\;\; \cdots \; (2)$
Substituting the values of $a^2$ and $b^2$ in equation $(2)$, we have,
$y = mx + \sqrt{3m^2 - 2}$ $\;\;\; \cdots \; (3)$
Equation $(3)$ passes through the point $\left(1, 2\right)$.
$\therefore$ $\;$ We have from equation $(3)$,
$2 = m + \sqrt{3m^2 - 2}$
i.e. $\;$ $2 - m = \sqrt{3m^2 - 2}$ $\;\;\; \cdots \; (4)$
Squaring both sides of equation $(4)$ we get,
$4 + m^2 - 4m = 3m^2 - 2$
i.e. $\;$ $2m^2 + 4m - 6 = 0$
i.e. $\;$ $m^2 + 2m - 3 = 0$
i.e. $\;$ $\left(m + 3\right) \left(m - 1\right) = 0$
i.e. $\;$ $m = -3$ $\;$ or $\;$ $m = 1$
Substituting $m = -3$ in equation $(3)$ gives
$y = -3x + \sqrt{3 \times \left(-3\right)^2 - 2}$
i.e. $\;$ $3x + y - 5 = 0$ $\;\;\; \cdots \; (5a)$
Substituting $m = 1$ in equation $(3)$ gives
$y = x + \sqrt{3 \times \left(1\right)^2 - 2}$
i.e. $\;$ $x - y + 1 = 0$ $\;\;\; \cdots \; (5b)$
Equations $(5a)$ and $(5b)$ are the required equations of tangents.