InterviewSolution
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InterviewSolution
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What is the product of the perpendiculars from the two points `(pm sqrt(b^(2)-a^(2)), 0)` to the line `ax cos phi + by sin phi =ab` ?A. `a^(2)`B. `b^(2)`C. `ab`D. `a//b` |
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Answer» Correct Answer - A Given equation of line is `ax cos phi +by sin phi -ab =0` Let `d_(1)` be the perpendicular distance from `(sqrt(b^(2)-a^(2)), 0)` to the line `ax cos phi + by sin phi-ab=0` and `d_(2)` from `(-sqrt(b^(2)-a^(2)), 0)` to the line `ax cos phi + by sin phi -ab =0` At point `(sqrt(b^(2)-a^(2)), 0)` `d_(1)=(asqrt(b^(2)-a^(2)) cos phi -ab)/sqrt(a^(2) cos^(2) phi +b^(2) sin^(2) phi)` At point `(-sqrt(b^(2)-a^(2)), 0)` `d_(2)=(-asqrt(b^(2)-a^(2))cos phi-ab)/sqrt(a^(2) cos^(2) phi +b^(2) sin^(2) phi)` `:. d_(1)d_(2)=-([a^(2)(b^(2)-a^(2)) cos^(2) phi-a^(2)b^(2)])/(a^(2) cos^(2) phi+b^(2) sin^(2) phi)` `= - (a^(2) (-b^(2) sin^(2) phi -a^(2) cos^(2) phi))/(a^(2) cos^(2) phi+b^(2) sin^(2) phi)=a^(2)` |
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