InterviewSolution
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InterviewSolution
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Provethat the sum of eccentric angles of four concylic points on the ellipse(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is 2npi, wheren in Z |
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Answer» Solution :Let the CIRCLE `x^(2)+y^(2)+2gx+2fy+e-0` CUT the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` in four ponts , P,Q,R and S. Solving circle and ellipse `(x=a cos theta, y=b sin theta)`, we have `a^(2)cos^(2)theta+b^(2)theta+2agcos theta+2bf sin theta+c=0` `rArra^(2)((1-t^(2))/(1+t))+b^(2)((2t)/(1+r^(2)))^(2)+2ag((1-r^(2))/(1+t^(2)))+2bf((2t)/(1+t))+c=0"where"t=tan.(theta)/(2)` `rArra^(2)(1-r^(2))^(2)+4b^(2)t^(2)+2ag(1-t^(2))^(2)(1-t^(2))+4bft(1+f^(2))+c(1+r^(2))^(2)=0` `rArr(a^(2)-2ag+c)t^(4)+4bft^(3)+(-2a^(2)+4b^(2)+2c)t^(2)+4bft+(a^(2)+2ag+c)-0` Roots of equation are `tan.(alpha)/(2),tan.(beta)/(2),tan.(gamma)/(2) and tan. (delta)/(2)`, where `alpha,beta,gamma and delta` ECCENTRIC angles of P,Q,R and S respectively . Now,`tan ((alpha)/(2)+(beta)/(2)+(gamma)/(2)+(delta)/(2))=(s_(1)-s_(2))/(1-s_(2)+s_(4))` Wherem `s_(i)` = sum of products of tangents of HALF angles taken 'i' at a time CLEARLY, from eqution (1), `s_(1)=s_(3)=(-4bf)/(a^(2)-2ag+c)` `tan ((alpha)/(2)+(beta)/(2)+(gamma)/(2)+(delta)/(2))=0` `rArr(alpha+beta+gamma+delta)/(2)=npi,n in Z` `rArr alpha+beta+gamma+delta=2npi,n in Z` |
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