Correct option is (D) \(\sum\) sec \(\alpha\) . \(\sum\) cos \(\alpha\) = 4
If foot of the normal of ellipse is (a cos θ, b sin θ) then the equation of the normal is
ax sec θ- by cosec θ = p2 - q2----(1)
where θ is eccentric angle of arbitary point on ellipse.
\(\therefore\) Let if equation(1) passes through point (h, k) then
ah sec θ - bk cosec θ = p2- q2
Let A = ah, B = p2 - p2, C = bk
Then A sec θ - C cosec θ = B
⇒ A sec θ - B = C cosec θ
⇒ A2 sec2 θ - 2AB sec θ + B2 = C2 cosec2 θ
(By squaring on bothsides)
⇒ A2 sec2θ - 2AB sec θ + B2 = \(\frac{C^2sec^2\theta}{sec^2\theta-1}\) \(\left(\because cosec^2\theta=\frac1{sin^2\theta}=\frac1{1-cos^2\theta}=\frac1{1-\cfrac1{sec^2\theta}}=\frac{sec^2\theta}{sec^2\theta-1}\right)\)
⇒ A2 sec4 θ - 2AB sec3 θ + B2 sec2θ - A2sec2θ + 2 AB sec θ - B2 = C2sec2θ
⇒ A2 sec4θ - 2AB sec3θ + (B2 - A2 - C2) sec2θ + 2AB secθ - B2 = 0---(2)
\(\because\) α, ß, \(\gamma\) & \(\delta\) are eccentric angle of point A, B, C and D and normals at points A, B, C & D are concurrent.
Let they are passing through (h, k)
Then α, ß, \(\gamma\) & \(\delta\) are roots of equation (2)
\(\therefore\) \(\sum\)sec θ = \(\frac{-(-2AB)}{A^2}\) (Sum of roots of equation(2))
⇒ \(\sum\)sec θ = \(\frac{2B}A\)---(3)
Now, \(\sum\) cos θ = \(\sum\)\(\frac1{sec\theta}\)
= \(\frac1{sec\alpha}+\frac1{sec\beta}+\frac1{sec\gamma}+\frac1{sec\delta}\)
= \(\frac{sec\beta.sec\gamma.sec\delta+sec\alpha.sec\gamma.sec\delta+sec\alpha.sec\beta.sec\delta+sec\alpha.sec\beta.sec\gamma}{sec\alpha.sec\beta.sec\gamma.sec\delta}\)
= \(\frac{\sum sec\alpha.sec\beta.sec\gamma}{\pi sec\alpha}\)
= \(\cfrac{\frac{-2AB}{A^2}}{\frac{-B^2}{A^2}}\) = \(\frac{2AB}{B^2}=\frac{2A}B\)
\(\therefore\) \(\sum\) cos θ = \(\frac{2A}{B}\)---(4)
From (3) & (4), we get
\(\sum\) sec θ .\(\sum\) cos θ = \(\frac{2B}A.\frac{2A}B\) = 4
or \(\sum\) sec \(\alpha\) . \(\sum\) cos \(\alpha\) = 4