Question

Let `P` be any point on a directrix of an ellipse of eccentricity `e ,S` be the corresponding focus, and `C` the center of the ellipse. The line `P C` meets the ellipse at `Adot` The angle between `P S` and tangent a `A` is `alpha` . Then `alpha` is equal to `tan^(-1)e` (b) `pi/2` `tan^(-1)(1-e^2)` (d) none of these
A. `tan^(-1)e`
B. `pi//2`
C. `tan^(-1)(1-e^(2))`
D. none of these

Answer 1

Let the ellipse be
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
and let `A-=(a cos theta, b sin theta)`
The equation of AC will be
`y=(b)/(a) tan thetax`
Solving with x=a/e, we get
`P-=((a)/(e)(b)/(e) tan theta)`
Slop of tangent at `A=-(b)/(a tan theta)`
Slope of `PS=((b)/(e) tantheta)/((a)/(e)-ae)=(b tantheta)/(a(1-e^(2)))=(a)/(b)tan theta`
So, `alpha=(pi)/(2)`