Conic Sections Question 266
Question: The line $ lx+my+n=0 $ is a normal to the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ , if
[DCE 2000]
Options:
A) $ \frac{a^{2}}{m^{2}}+\frac{b^{2}}{l^{2}}=\frac{(a^{2}-b^{2})}{n^{2}} $
B) $ \frac{a^{2}}{l^{2}}+\frac{b^{2}}{m^{2}}=\frac{{{(a^{2}-b^{2})}^{2}}}{n^{2}} $
C) $ \frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{{{(a^{2}-b^{2})}^{2}}}{n^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
The equation of any normal to $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ is $ ax\sec \theta - $ by $ cosec\theta =a^{2}-b^{2} $ ……(i) The straight line $ lx+my+n=0 $ ……(ii) will be a normal to the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $
If (i) and (ii) represent the same line then, $ \frac{a\sec \theta }{l}=\frac{bcosec\theta }{-m}=\frac{a^{2}-b^{2}}{-n} $
$ \Rightarrow \cos \theta =\frac{-an}{l(a^{2}-b^{2})} $ and $ \sin \theta =\frac{bn}{m(a^{2}-b^{2})} $
$ \because $
$ {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 $
$ \frac{a^{2}n^{2}}{l^{2}{{(a^{2}-b^{2})}^{2}}}+\frac{b^{2}n^{2}}{m{{(a^{2}-b^{2})}^{2}}}=1 $
therefore $ \frac{a^{2}}{l^{2}}+\frac{b^{2}}{m^{2}}=\frac{{{(a^{2}-b^{2})}^{2}}}{n^{2}} $ .
BETA
