Find the equation of tangents to the hyperbola $4x^2 - y^2 = 64$, which are parallel to $10x-3y+9=0$.
Equation of hyperbola is $\;$ $4x^2 - y^2 = 64$
i.e. $\;$ $\dfrac{x^2}{64 / 4} - \dfrac{y^2}{64} = 1$
i.e. $\;$ $\dfrac{x^2}{16} - \dfrac{y^2}{64} = 1$ $\;\;\; \cdots \; (1)$
The transverse axis of the hyperbola given by equation $(1)$ is along the X axis.
Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives
$a^2 = 16 \implies a = 4$ $\;$ and $\;$ $b^2 = 64 \implies b = 8$
Equation of given line is $\;$ $10x - 3y + 9 = 0$
i.e. $\;$ $y = \dfrac{10}{3} x + 3$
Slope of given line $= m = \dfrac{10}{3}$
$\because$ $\;$ the required tangents are parallel to the given line,
$\therefore$ $\;$ slope of required tangents $= m = \dfrac{10}{3}$
Equations of tangents (with slope $m$) to the hyperbola are $\;$ $y = mx \pm \sqrt{a^2 m^2 - b^2}$
$\therefore$ $\;$ Required equations of tangents to the hyperbola are
$y = \dfrac{10}{3} x \pm \sqrt{16 \times \dfrac{100}{9} - 64}$
i.e. $\;$ $y = \dfrac{10}{3} x \pm \sqrt{\dfrac{1024}{9}}$
i.e. $\;$ $y = \dfrac{10}{3} x \pm \dfrac{32}{3}$
i.e. $\;$ $10 x - 3y \pm 32 = 0$