Question: The product of the perpendiculars drawn from the points $ (\pm \sqrt{a^{2}-b^{2},}0) $ on the line $ \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1 $ , is
Options:
A) $ a^{2} $
B) $ b^{2} $
C) $ a^{2}+b^{2} $
D) $ a^{2}-b^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ ( \frac{b\sqrt{a^{2}-b^{2}}\cos \theta +0-ab}{\sqrt{b^{2}{{\cos }^{2}}\theta +a^{2}{{\sin }^{2}}\theta }} )( \frac{-b\sqrt{a^{2}-b^{2}}\cos \theta -ab}{\sqrt{b^{2}{{\cos }^{2}}\theta +a^{2}{{\sin }^{2}}\theta }} ) $ $ =\frac{-[b^{2}(a^{2}-b^{2}){{\cos }^{2}}\theta -a^{2}b^{2}]}{(b^{2}{{\cos }^{2}}\theta +a^{2}{{\sin }^{2}}\theta )} $ $ =\frac{b^{2}[a^{2}-a^{2}{{\cos }^{2}}\theta +b^{2}{{\cos }^{2}}\theta ]}{b^{2}{{\cos }^{2}}\theta +a^{2}{{\sin }^{2}}\theta } $ $ =\frac{b^{2}[a^{2}{{\sin }^{2}}\theta +b^{2}{{\cos }^{2}}\theta ]}{b^{2}{{\cos }^{2}}\theta +a^{2}{{\sin }^{2}}\theta } $ = $ b^{2} $ .Trick: Let $ a=2,b=1 $ and $ \theta =\frac{\pi }{2} $ , then the points are $ (\pm \sqrt{3},0) $ and the line is y = 1. Length from $ (\sqrt{3},0) $ on $ y=1 $ is 1 and that of from $ (-\sqrt{3},0) $ is also 1. Hence product is $ 1\times 1=1 $ , which is given by (b).
BETA
