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 $$\frac{{24}}{{{m^2}}} - \frac{{18}}{{{{\left( { - 1} \right)}^2}}} = \frac{{{{\left( {24 + 18} \right)}^2}}}{{{{\left( {7\sqrt 3 } \right)}^2}}}\,\, \Rightarrow m = \frac{2}{{\sqrt 5 }}$$