Conic Sections Question 253
Question: If the line $ x\cos \alpha +y\sin \alpha =p $ be normal to the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ , then
[MP PET 2001]
Options:
A) $ p^{2}(a^{2}{{\cos }^{2}}\alpha +b^{2}{{\sin }^{2}}\alpha )=a^{2}-b^{2} $
B) $ p^{2}(a^{2}{{\cos }^{2}}\alpha +b^{2}{{\sin }^{2}}\alpha )={{(a^{2}-b^{2})}^{2}} $
C) $ p^{2}(a^{2}{{\sec }^{2}}\alpha +b^{2}cose{c^{2}}\alpha )=a^{2}-b^{2} $
D) $ p^{2}(a^{2}{{\sec }^{2}}\alpha +b^{2}cose{c^{2}}\alpha )={{(a^{2}-b^{2})}^{2}} $
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Answer:
Correct Answer: D
Solution:
The equation of any normal to $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ is $ ax\sec \varphi -bycosec\varphi =a^{2}-b^{2} $ ……(i) The straight line $ x\cos \alpha +y\sin \alpha =p $ will be a normal to the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $
If (i) and $ x\cos \alpha +y\sin \alpha =p $ represent the same line $ \frac{a\sec \varphi }{\cos \alpha }=\frac{-bcosec\varphi }{\sin \alpha }=\frac{a^{2}-b^{2}}{p} $
$ \Rightarrow \cos \varphi =\frac{ap}{(a^{2}-b^{2})\cos \alpha }, $
$ \sin \varphi =\frac{-bp}{(a^{2}-b^{2})\sin \alpha } $
$ \because $
$ {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi =1 $
$ \Rightarrow \frac{b^{2}p^{2}}{{{(a^{2}-b^{2})}^{2}}{{\sin }^{2}}\alpha }+\frac{a^{2}p^{2}}{{{(a^{2}-b^{2})}^{2}}{{\cos }^{2}}\alpha }=1 $
$ \Rightarrow $ $ p^{2}(b^{2}cose{c^{2}}\alpha +a^{2}{{\sec }^{2}}\alpha )={{(a^{2}-b^{2})}^{2}} $ .
BETA
