Official solution
Let foot of perpendicular is (h, k)
Given \[{{{x^2}} \over 4} + {{{y^2}} \over 2} = 1\]
\[ \therefore \] a = 2, b = \[\sqrt 2 \]
and e = \[\sqrt {1 - {2 \over 4}} \] = \[{1 \over {\sqrt 2 }}\]
\[ \therefore \] Focus(ae, 0) = \[\left( {\sqrt 2 ,0} \right)\]
Equation of tangent
y = mx + \[\sqrt {{a^2}{m^2} + {b^2}} \]
\[ \Rightarrow \] y = mx + \[\sqrt {4{m^2} + 2} \]
Passes through (h, k)
(k -- mh)2
= 4m2
+ 2 .....(1)
Line perpendicular to tangent will have slope \[ - {1 \over m}\].
y - 0 = \[ - {1 \over m}\left( {x - \sqrt 2 } \right)\]
my = -x + \[{\sqrt 2 }\]
\[ \therefore \] (h + mk)2 = 2 .....(2)
Add equation (1) and (2)
k2(1 + m2) + h2
(1 + m2) = 4 (1 + m2)
h2
+ k2
= 4
x2
+ y2
= 4 (Auxiliary circle)
\[ \therefore \] \[\left( { - 1,\sqrt 3 } \right)\] lies on the locus.