Conic Sections Question 417
Question: The value of m, for which the line $ y=mx+\frac{25\sqrt{3}}{3} $ , is a normal to the conic $ \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 $ , is
[MP PET 2004]
Options:
A) $ \sqrt{3} $
B) $ -\frac{2}{\sqrt{3}} $
C) $ -\frac{\sqrt{3}}{2} $
D) 1
Show Answer
Answer:
Correct Answer: B
Solution:
We know that the equation of the normal of the conic $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ at point $ (a\sec \theta ,b\tan \theta ) $ is $ ax\sec \theta +by\cot \theta =a^{2}+b^{2} $
or $ y=\frac{-a}{b}\sin \theta x+\frac{a^{2}+b^{2}}{b\cot \theta } $
Comparing above equation with equation $ y=mx+\frac{25\sqrt{3}}{3} $ and taking $ a=4,b=3 $
we get,
$ \frac{a^{2}+b^{2}}{b\cot \theta }=\frac{25\sqrt{3}}{3} $
$ \Rightarrow $ $ \tan \theta =\sqrt{3}\Rightarrow \theta =60^{o} $
and $ m=-\frac{a}{b}\sin \theta =\frac{-4}{3}\sin 60^{o} $ = $ \frac{-4}{3}\times \frac{\sqrt{3}}{2}=\frac{-2}{\sqrt{3}} $ .
BETA
