Conic Sections Question 293
Question: The value of $ \lambda $ , for which the line $ 2x-\frac{8}{3}\lambda y=-3 $ is a normal to the conic $ x^{2}+\frac{y^{2}}{4}=1 $ is
[MP PET 2004]
Options:
A) $ \frac{\sqrt{3}}{2} $
B) $ \frac{1}{2} $
C) $ -\frac{\sqrt{3}}{2} $
D) $ \frac{3}{8} $
Show Answer
Answer:
Correct Answer: D
Solution:
We know that the equation of the normal at point $ (a\cos \theta ,b\sin \theta ) $ on the curve $ x^{2}+\frac{y^{2}}{4}=1 $ is given by $ ax\sin \theta -by\text{cosec }\theta =a^{2}-b^{2} $ ……(i) Comparing equation (i) with $ 2x-\frac{8}{3}\lambda y=-3 $ . we get,
$ a\sin \theta =2 $ , $ b\text{ cosec}\theta =\frac{8}{3}\lambda $ or $ ab=\frac{16}{3}\lambda $ …..(ii) $ \because a=1,b=2 $ ;$ 2=\frac{16}{3}\lambda $ or $ \lambda =3/8 $
BETA
