1.

If alpha, beta, gamma are the eccentric angles of three points on the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 at which the normals are concurrent, then sin(alpha+beta)+sin(beta+gamma)+sin(gamma+alpha) is equal to______

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Solution :SUPPOSE normals are `alpha, beta` and `GAMMA` are concurrent at `(h,k)` and LET `S` be the foot the fourth normals from `(h,k)` then we have `sum "TAN"(alpha)/2 "tan"(beta)/2=0` and `"tan"(alpha)/2 "tan" (beta)2"tan"(gamma)/2 "tan" (delta)/2=-1`
Eliminating `"tan"(delt)/2` from above, we will get `SIN(alpha+beta)+sin(beta+gamma)+sin(gamma+alpha)=0`


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