Let L=logE8 where E is the value of eccentricity of the curve: tan100x+tan200y+tan300tan1200x+tan2200y+tan3200+2020=0 The value of L is
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answer is 6.
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Detailed Solution
m1m2=tan100tan200×tan1200tan2200=−3tan800tan200tan400=−3tan600=−1Hence the equation is rectangular hyperbola with eccentricity 2 since the given equation represents a hyperbola as ∆=abc +2fgh-af2-bg2-ch2≠0The two lines which are given in the parentheses of the given equation are asymptotes of the hyperbolaTherefore, log28=6