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If `p_(1) and p_(2)` are the lengths of the perpendicular form the orgin to the line `x sec theta+y cosec theta=a and xcostheta-y sin theta=a cos 2 theta` respectively then prove that `4p_(1)^(2)+p_(2)^(2)=a^(2)` |
Answer» We have `x sectheta+y "cosec" theta =a Rightarrow (x)/(cos theta)+(y)/(sin theta)-a=0.......(i)` Now, `p_(1)` is the length of perpendicular from the origin to line (i), so we have `p_(1)=(|(1)/(cos theta)xx0+(1)/(sin theta)xx0-a|)/(sqrt((1)/(cos^(2)theta)+(1)/(sin^(2)theta)))=|a|sin theta cos theta=(|a|)/(2)sin 2theta` `Rightarrow 2p_(1)=|a|sin 2theta` `Rightarrow 4p_(1)^(2)=a^(2)sin^(2) 2theta........(ii)` The other line is `x cos theta-y sin theta-a cos 2theta=0`.......(ii) Now, `p_(2)` is the length of perpendicular from the origin to line (ii), so we have `P_(2)=(|0xx cos theata-0xxsin theta-acos2 theta|)/(sqrt(cos^(2)theta+sin^(2)theta))=|a cos 2theta|` `Rightarrow p_(2)^(2)=a^(2) cos^(2)2 theta...(iv)` Adding (ii) and (iv), we get `4p_(1)^(2)+p_(2)^(2)=a^(2)(sin^(2)2theta+cos^(2)2theta)` `Rightarrow 4p_(1)^(2)+p_(2)^(2)=a^(2)` |
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