Question
Let $$P\left( {6,\,3} \right)$$ be a point on the hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1.$$ If the normal at the point $$P$$ intersects the $$x$$-axis at $$\left( {9,\,0} \right),$$ then the eccentricity of the hyperbola is :
A.
$$\sqrt {\frac{5}{2}} $$
B.
$$\sqrt {\frac{3}{2}} $$
C.
$$\sqrt 2 $$
D.
$$\sqrt 3 $$
Answer :
$$\sqrt {\frac{3}{2}} $$
Solution :
For hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1,$$ we have
$$\frac{{2x}}{{{a^2}}} - \frac{{2y}}{{{b^2}}}\frac{{dy}}{{dx}} = 0\,\,\, \Rightarrow \frac{{dy}}{{dx}} = \frac{{{b^2}x}}{{{a^2}y}}$$
$$\therefore $$ Slope of normal at $$P\left( {6,\,3} \right)$$
$$ = - \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\left( {6,\,3} \right)}}}} = - \frac{{3{a^2}}}{{6{b^2}}}$$
$$\therefore $$ Equation of normal is $$\frac{{y - 3}}{{x - 6}} = - \frac{{3{a^2}}}{{6{b^2}}}$$
As it intersects $$x$$-axis at $$\left( {9,\,0} \right)$$
$$\therefore \frac{{0 - 3}}{{9 - 6}} = \frac{{ - 3{a^2}}}{{6{b^2}}}\,\,\, \Rightarrow {a^2} = 2{b^2}.....(1)$$
Also for hyperbola, $${b^2} = {a^2}\left( {{e^2} - 1} \right)$$
Using $${a^2} = 2{b^2};$$ we get
$$\eqalign{
& {b^2} = 2{b^2}\left( {{e^2} - 1} \right) \cr
& \frac{1}{2} = {e^2} - 1 \cr
& {\text{or,}}\,\,{e^2} = \frac{3}{2} \cr
& {\text{or,}}\,\,e = \sqrt {\frac{3}{2}} \cr} $$