about to only mathematics

Options

A. `(a^(2)+b^(2))/(a)`
B. `-((a^(2)+b^(2))/(a))`
C. `(a^(2)+b^(2))/(b)`
D. `-((a^(2)+b^(2))/(b))`

Answer

D

Explanation

Normals at `P(theta)` and `Q(theta)` are
`ax cos theta+" by "cot theta=a^(2)+b^(2)`
`ax cos phi + " by " cot phi =a^(2)+b^(2)`
where `phi=(pi)/(2)-theta`
These normals pass through (h, k). Therefore,
`ah cos theta+bk cot theta=a^(2)+b^(2)`
and `ah sin theta+bk tan theta=a^(2)+b^(2)`
Eliminating h, we have
`bk(cot theta sin theta - tan theta cos theta)=(a^(2)+b^(2))(sin theta- cos theta)`
`"or "k=-((a^(2)+b^(2))/(b))`