Distance between feet of perpendiculars from focii of ellipse

Question

$$Tangent\quad drawn\quad to\quad ellipse\quad { x }^{ 2 }+{ 2y }^{ 2 }=6\quad at\quad point\quad (2,1).If\quad A\\ and\quad B\quad are\quad the\quad feet\quad of\quad pependiculars\quad from\quad the\quad two\\ focii\quad on\quad the\quad tangent,then\quad length\quad AB\quad equals\quad ? $$

Took this problem from https://brilliant.org/problems/are-you-smarter-than-me-13/

I solved the problem by finding feet of perpendiculars and then used distance formula.But is there an easier way?

Answer 1

The foci of this ellipse are $(\pm \sqrt 3,0)$ so the distance between them is $d=2\sqrt 3$.

The slope of the tangent T at $(2,1)$ is $m=-1$.

The required distance is simply the projection of segment $d$ onto the line T. Hence the required distance is $2\sqrt 3\div\sqrt 2=\sqrt 6$