If any two chords be drawn through two points on the major axis of an ellipse equidistant from the centre, show that `tan(alpha/2)* tan(beta/2) *tan(gamma/2)*tan(delta/2)=1`, where `alpha, beta, gamma, delta` are the eccentric angles of the .extremities of the chords.

Options

A. `-1`
B. `1`
C. `(a)/(b)`
D. `(b)/(a)`

Answer

B

Explanation

let PQ and RS be two chords of the ellipse such that the eccentric angles of P,Q R and S ` alpha , beta gamma and delta ` respectively .
suppose PQ and RS meet major axis at `(c,0) and (-c,0)` respectivley , then
`tan ""(alpha)/(a) tan ""(beta)/(2) =(c-a)/(c+a)` [ see illustration 1]
`and tan "" ( gamma)/(2) tan ""( delta)/(2)=(-c-a)/(-c+a)=(c+a)/(c-a)`
` therefore tan "" ( alpha )/(2) tan "" ( beta)/(2) tan ""( gamma)/(2)tan ""( delta )/(2) =(c-a)/(c+a)xx(c+a)/(c-a)=1`