Applications-Of-Derivatives Question 477
Question: The normal to the curve $ x=a(1+\cos \theta ),,y=a\sin \theta $ at $ ‘\theta ’ $ always passes through the fixed point
[AIEEE 2004]
Options:
A) (a, a)
B) (0, a)
C) (0, 0)
D) (a, 0)
Show Answer
Answer:
Correct Answer: D
Solution:
Slope of normal $ =\frac{-dx}{dy}=\frac{\frac{-d{ a(1+\cos \theta ) }}{d\theta }}{\frac{d(a\sin \theta )}{d\theta }}=\frac{a\sin \theta }{a\cos \theta }=\tan \theta $ Now, the equation of normal at $ \theta $ is, $ y-a\sin \theta =\tan \theta [x-a(1+\cos \theta )] $ Clearly, this line passes through (a, 0).
BETA
