Applications Of Derivatives Question 135
Question: The lines tangent to the curves $ y^{3}-x^{2}y+5y-2x=0 $ and $ x^{4}-x^{3}y^{2}+5x+2y=0 $ at the origin intersect at an angles $ \theta $ equal to
Options:
A) $ \frac{\pi }{6} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{3} $
D) $ \frac{\pi }{2} $
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Answer:
Correct Answer: D
Solution:
[d] Differentiating $ y^{3}-x^{2}y+5y-2x=0 $ w.r.t. x, we get $ 3y^{2}y’-2xy-x^{2}y’+5y’-2=0 $ Or $ y’=\frac{2xy+2}{3y^{2}-x^{2}+5} $ or $ y{’ _{(0,0)}}=\frac{2}{5} $
Differentiating $ x^{4}-x^{3}y^{2}+5x+2y=0 $ w.r.t. x, we get $ 4x^{2}-3x^{2}y^{2}-2x^{3}yy’+5+2y’=0 $ Or $ y’=\frac{3x^{2}y^{2}-4x^{3}-5}{2-2x^{3}y} $ or $ y{’ _{(0,0)}}=-\frac{5}{2}. $
Thus, both the curves intersect at right angle.
BETA
