If PSQ + PHR be two chords of an ellipse through its foci S and H, then prove that `(PS)/(SQ) + (PH)/(HQ) = 2(1 + e^2)/(1-e^2)`, Where e is the eccentricity of ellipse.

If PSQ + PHR be two chords of an ellipse through its foci S and H, the

1.	If PSQ + PHR be two chords of an ellipse through its foci S and H, then prove that `(PS)/(SQ) + (PH)/(HQ) = 2(1 + e^2)/(1-e^2)`, Where e is the eccentricity of ellipse.
Answer» We can draw an ellipse with the given details. Please refer to video for the figure. Equation of ellipse with respect to focus `S` as a pole, `l/r = 1+ecos theta` Here, `r = SP`, `:. l/(SP) = 1+ e cos theta ->(1)` If vectorial angle of `P` is `theta`, then vectorial angle of `Q` will be `pi+theta`. If we write equation with respect to `SQ`, then, `:. l/(SQ) = 1+ ecos(pi+theta)` `=> l/(SQ) = 1-ecos theta->(2)` Adding (1) and (2), `l/(SP)+l/(SQ) = 2` `=> 1/(SP)+1/(SQ) = 2/l` `=>1+(SP)/(SQ) = (2(SP))/l->(3)` Similarly, for chord `PHR`, we can have, `1+(PH)/(HR) = (2(PH))/l->(4)` Adding (3) and (4), `2+((SP)/(SQ)+(PH)/(HR)) = 2/l(SP+PH)->(5)` As, `S` and `H` are foci of given ellipse.>br>`:. SP+PH =` length of major axis `= 2a` Also, we know, `2l = 2b^2/a => l =b^2/a` Putting these values in equation (5), `2+((SP)/(SQ)+(PH)/(HR)) = 4a^2/b^2` Als, we know, `b^2 = a^2(1-e^2)` So, our equation beecomes, `((SP)/(SQ)+(PH)/(HR)) = 4a^2/(a^2(1-e^2))-2` `=>((SP)/(SQ)+(PH)/(HR)) = (2+2e^2)/(1-e^2)` `=>((SP)/(SQ)+(PH)/(HR)) = (2(1+e^2))/(1-e^2)`