Conic Sections Question 344
Question: The condition that the straight line $ lx+my=n $ may be a normal to the hyperbola $ b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2} $ is given by
[MP PET 1993, 94]
Options:
A) $ \frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{{{(a^{2}+b^{2})}^{2}}}{n^{2}} $
B) $ \frac{l^{2}}{a^{2}}-\frac{m^{2}}{b^{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) $ \frac{l^{2}}{a^{2}}+\frac{m^{2}}{b^{2}}=\frac{{{(a^{2}-b^{2})}^{2}}}{n^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Any normal to the hyperbola is $ \frac{ax}{\sec \theta }+\frac{by}{\tan \theta }=a^{2}+b^{2} $ ……(i) But it is given by $ lx+my-n=0 $ ……(ii) Comparing (i) and (ii), we get $ \sec \theta =\frac{a}{l}( \frac{-n}{a^{2}+b^{2}} ) $ and $ \tan \theta =\frac{b}{m}( \frac{-n}{a^{2}+b^{2}} ) $
Hence eliminating $ \theta $ , we get $ \frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{{{(a^{2}+b^{2})}^{2}}}{n^{2}} $ .
BETA
