Question

If the normal at the four points \( A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right) \& D\left(x_{4}, y_{4}\right) \) on the ellipse \( \frac{x^{2}}{p^{2}}+\frac{y^{2}}{q^{2}}=1 \) are concurrent. The eccentric angles of \( A, B, C, D \) are \( \alpha, \beta, \gamma \& \delta \) respectively. If \( x_{1}, x_{2}, x_{3}, x_{4} \), are roots of the equation \( a x^{4}+b x^{3}+c x^{2}+d x+e=0 \). Then 

(A) \( b d=4 a e \) 

(B) \( b d=3 a e \) 

(C) \( \sum(\cos \alpha) \cdot \sum(\sec \alpha)=3 \) 

(D) \( \sum(\cos \alpha) \cdot \sum(\sec \alpha)=4 \)

Answer 1

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