Questions
are the eccentric angles of the extremities of a focal chord of the ellipse then
a
7+47−4
b
-923
c
5−45+4
d
87−239
detailed solution
Correct option is D
The eccentricity e=1−916=74Let P(4cosα,3sinα) and Q(4cosβ,3sinβ) be a focalchord of the ellipse passing through the focus at (7,0)then,⇒3sinβ4cosβ−7=3sinα4cosα−7⇒sin(α−β)sinα−sinβ=74⇒cos[(α−β)/2]cos[(α+β)/2]=74⇒tanα2tanβ2=7−47+4=23−87−9Talk to our academic expert!
Similar Questions
If a, b are the eccentric angles of the
extremities of a focal chord of the ellipse
then
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