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If the chord joining points `P(alpha) and Q(beta)` on the ellipse `((x^2)/(a^2))+((y^2)/(b^2))=1` subtends a right angle at the vertex `A(a ,0),` then prove that `tan(a/2)tan(beta/2)=-(b^2)/(a^2)dot`

Text Solution

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`m_(PA)xxm_(QA)=-1`
`or (b sin alpha)/(a cos alpha-a)xx(b sin beta)/(a cos beta-a)=1`
`or (4 sin .(alpha)/(2)cos.(alpha)/(2)sin.(beta)/(2)cos.(beta)/(2))/((2sin^(2).(alpha)/(2))(2sin^(2).(beta)/(2)))=(-a^(2))/(b^(2))`

or `tan.(alpha)/(2)tan.(beta)/(2)=-(b^(2))/(a^(2))`
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