Coordinate geometry and Trignometry. Find the condition so that

Pizzadililehz

Pizzadililehz

Answered question

2022-03-27

Coordinate geometry and Trignometry.
Find the condition so that the line px+qy=r intersects the ellipse x2a2+y2b2=1 in points whose eccentric angles differ by π4.
Though I know how to solve it using parametric coordinates, I was wondering if there's an another approach which is less time consuming.

Answer & Explanation

Alannah Farmer

Alannah Farmer

Beginner2022-03-28Added 11 answers

Let P1=(acost,bsint) and P2=(acos(t+ϕ),bsin(t+ϕ)) where ϕ=dπ4, then we need to satisfy two conditions:nSK1. P1 is on the line px+qy=r, therefore
pacost+qbsint=r (1)
2. P2 is on the line as well, hence,
pacos(t+ϕ)+qbsin(t+ϕ)=r (2)
This last equation becomes
cost(pacosϕ+qbsinϕ)+sint(-pasinϕ+qbcosϕ)=r (3)
And after substituting cosϕ=sinϕ=d12, it becomes,
cost(pa+qb)+sint(qb-pa)=2r (4)
Now solving (1) and (4) for cost and sint, we find that
cost=rd(qbpa)2qbp2a2+q2b2
sint=rd2pa(pa+qb)p2a2+q2b2
Since cos2t+sin2t=1, then the condition is
d(p2a2+q2b2)2r2=(qbpa)2+(pa+qb)2+2(q2b2+p2a2)22(qb(qbpa)+pa(pa+qb))
Simplifying the right hand side,
d(p2a2+q2b2)2r2=4(p2a2+q2b2)22(q2b2+p2a2)
Dividing through by (p2a2+q2b2), yields,
dp2a2+q2b2r2=422
And this is the condition on the line.

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