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Question

If line lx+my+n=0 cuts the ellipse x2a2+y2b2=1 at points whose eccentric angles differ by π2, then the value of a2l2+b2m2n2 is

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Solution

Let eccentric angles are θ and ϕ, then
θϕ=π2 (given)
θ=π2+ϕ
The line joining the points θ and ϕ is
xacos(θ+ϕ2)+ybsin(θ+ϕ2)=cos(θϕ2)
(θ=π2+ϕ)
xacos(π4+ϕ)+ybsin(π4+ϕ)=cosπ4
xacos(π4+ϕ)+ybsin(π4+ϕ)=12 ...(1)

The given line is lx+my+n=0
lx+my=n ...(2)

Now, equation (1) and (2) represent the same line, so comparing them, we get
cos(π4+ϕ)la=sin(π4+ϕ)mb=1n2
cos(π4+ϕ)=lan2...(3)
and
sin(π4+ϕ)=mbn2...(4)
Squaring and adding (3) and (4), we get
l2a22n2+m2b22n2=1
l2a2+m2b2=2n2
a2l2+b2m2n2=2

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