Question

If the line $$y = mx - 7\sqrt 3 $$    is normal to the hyperbola $$\frac{{{x^2}}}{{24}} - \frac{{{y^2}}}{{18}} = 1,$$   then a value of $$m$$ is :

A. $$\frac{{\sqrt {5} }}{2}$$
B. $$\frac{{\sqrt {15} }}{2}$$
C. $$\frac{2}{{\sqrt 5 }}$$  
D. $$\frac{3}{{\sqrt 5 }}$$
Answer :   $$\frac{2}{{\sqrt 5 }}$$
Solution :
Since, $$lx + my + n = 0$$    is a normal to $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1,$$   then $$\frac{{{a^2}}}{{{l^2}}} - \frac{{{b^2}}}{{{m^2}}} = \frac{{{{\left( {{a^2} + {b^2}} \right)}^2}}}{{{n^2}}}$$     but it is given that $$mx - y + 7\sqrt 3 $$    is normal to hyperbola $$\frac{{{x^2}}}{{24}} - \frac{{{y^2}}}{{18}} = 1$$   then $$\frac{{24}}{{{m^2}}} - \frac{{18}}{{{{\left( { - 1} \right)}^2}}} = \frac{{{{\left( {24 + 18} \right)}^2}}}{{{{\left( {7\sqrt 3 } \right)}^2}}}\,\, \Rightarrow m = \frac{2}{{\sqrt 5 }}$$