If the line y = mx + 7 3 is normal to the hyperbola x ^ 2 24 - y ^...

The given equation of the hyperbola is $\dfrac{{x}^{2}}{24} - \dfrac{{y}^{2}}{18} = 1,$ which is of the form $\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1 .$

$⇒ {a}^{2} = 24; {b}^{2} = 18$

We know that the equation of a normal of slope $m$ to the hyperbola $\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1$ is $y = mx \pm \dfrac{m\left({a}^{2} + {b}^{2}\right)}{\sqrt{{a}^{2} - {b}^{2}{m}^{2}}}$

Thus, the equation of the normal to the given hyperbola is $y = mx \pm \dfrac{m\left(24 + 18\right)}{\sqrt{24 - 18{m}^{2}}}$

$y = mx \pm \dfrac{m\left(42\right)}{\sqrt{24 - 18{m}^{2}}}$

Given, the normal is $y = mx + 7\sqrt{3}$

Hence, comparing the two equations, we get $\dfrac{42m}{\sqrt{24 - 18{m}^{2}}} = 7\sqrt{3}$

$⇒ \dfrac{{\left(42m\right)}^{2}}{24 - 18{m}^{2}} = 49 \times 3$

$⇒ 42 \times 42{m}^{2} = 49 \times 3\left(24 - 18{m}^{2}\right)$

$⇒ 12{m}^{2} = 24 - 18{m}^{2}$

$⇒ {m}^{2} = \dfrac{24}{30} = \dfrac{4}{5}$

$⇒ m = \pm \sqrt{\dfrac{4}{5}} = \pm \dfrac{2}{\sqrt{5}} .$