Conic Challenge 10

This Question is all about " Symmetry "

Minimum Normal Must be Lie along the Line y=x , and again by symmetry length of Normal in 1st quadrant and 3rd quadrant will be equal .

Now Parametric Point's will be $\displaystyle{A(ct,\cfrac { c }{ t } )\quad and\quad B(-ct,-\cfrac { c }{ t } )}$

Length of Normal : $\displaystyle{A(ct,\cfrac { c }{ t } )\quad and\quad B(-ct,-\cfrac { c }{ t } )\\ L=AB=2\times OA=2\times OB\\ { L }^{ 2 }=4[{ c }^{ 2 }{ t }^{ 2 }+{ (\cfrac { c }{ t } })^{ 2 }]}$

Now Use AM-GM ::( At Minimum AM=GM and GM=constant and equality of variables must be attained )

$\boxed { { L }^{ 2 }=8{ c }^{ 2 } }$

Note that the minimum possible distance between the two branches of the hyperbola happen at the line segment that lies on the line $y=x$ . It just so happens that this line segment is also normal to the parabola, so all we need to do is find the length of the line segment.

The segment starts at the coordinate $(c,c)$ and ends at $(-c,-c)$ . Thus, the length is $\sqrt{(2c)^2+(2c)^2}=c\sqrt{8}$

Then $k=\sqrt{8}$ so $k^2=\boxed{8}$