Straight Line Question 207
Question: Locus of midpoint of the portion between the axes of $ x\cos \alpha +y\sin \alpha =p $ where p is constant is
Options:
A) $ x^{2}+y^{2}=\frac{4}{p^{2}} $
B) $ x^{2}+y^{2}=4p^{2} $
C) $ \frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{2}{p^{2}} $
D) $ \frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{4}{p^{2}} $
Correct Answer: D $ \Rightarrow \frac{x\cos \alpha }{p}+\frac{y\sin \alpha }{p}=1; $ $ \Rightarrow \frac{x}{p/\cos \alpha }+\frac{y}{p/\sin \alpha }=1 $
So co-ordinates of A and B are
$ ( \frac{p}{\cos \alpha },0 ) $ and $ ( 0,\frac{p}{\sin \alpha } ) $ ;
So co-ordinates of midpoint of AB are
$ ( \frac{p}{2\cos \alpha },\frac{p}{2\sin \alpha } )=(x_1,y_1)(say); $ $ x_1=\frac{p}{2\cos \alpha }\And y_1=\frac{p}{2\sin \alpha }; $ $ \Rightarrow \cos \alpha =p/2x_1 $ and $ \sin \alpha =p/2y_1; $
Consider $ {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha =1 $ $ \Rightarrow \frac{p^{2}}{4}( \frac{1}{x_1^{2}}+\frac{1}{y^2_1} )=1 $ $ \therefore $ Locus of $ (x_1,y_1) $ is $ \frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{4}{p^{2}}. $Show Answer
Answer:
Solution:
BETA
