1) If the normal at one end of the latus rectum of an ellipse x2/a2+y2/b2=1 passes through the one end of the minor axis then
A. e4-e2+1=0 B. e2-e+1=0
C. e2+e+1=0 D. e4+e2-1=0
2) In an ellipse the eccentric angle of any point P measured from the semi major axis CA is \phi. If S be the focous nearest to A and angle ASP=\theta, e=1/2 then [tan(\theta/2)] / [tan(\phi/2)]=?
29Ans 1 Eqn of normal at (ae, b2/a) ends of latus rectum
x-ey=ae -\frac{b^{2}}{a}e
it passes through (0,b)
so we get
\frac{b}{a} = \frac{b^{2}}{a^{2}} - 1
Now put
\frac{b^{2}}{a^{2}} = 1-e^{2}
and \frac{b}{a} =\sqrt{1-e^{2}}
Squaring both sides.. u will get the answer D
3tanθ=b sinφacosφ-ae
=(b/a) sinφcosφ-e
thus we get tanθ=√1-e2sinφcosφ-e
thus sinθcosφ-esinθ=√1-e2sinφ cosθ
put sinθ in terms of tan(θ/2), cosφ terms of terms of tan(θ/2)---
simplifying we get
(√1+etan(φ/2)+√1-ecot(θ/2) ) (√1+etan(φ/2)-√1-etan(θ/2) )=0
but θ and φ lies between 0 and Î
tan(θ/2)>0 and tan(φ/2)>0
(√1+etan(φ/2)+√1+ecot(θ/2) ) ≠0
but (√1+etan(φ/2)-√1+etan(θ/2) )=0
thus [tan(θ/2)] / [tan(φ/2)= √1+e√1-e
ans -------√3