Applications Of Derivatives Question 159
Question: The equation of the tangent to the curve $ x=2{{\cos }^{3}}\theta $ and $ y=3{{\sin }^{3}}\theta $ at the point $ \theta =\pi /4 $ is
[J & K 2005]
Options:
A) $ 2x+3y=3\sqrt{2} $
B) $ 2x-3y=3\sqrt{2} $
C) $ 3x+2y=3\sqrt{2} $
D) $ 3x-2y=3\sqrt{2} $
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Answer:
Correct Answer: C
Solution:
$ {{. x, |} _{\theta =\frac{\pi }{4}}}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}, $
$ {{. y, |} _{\theta =\frac{\pi }{4}}}=\frac{3}{2\sqrt{2}},{{. \frac{dy}{dx} |} _{\theta =\frac{\pi }{4}}}{{. \frac{9,{{\sin }^{2}}\theta \cos \theta }{-6,{{\cos }^{2}}\theta \sin \theta } |} _{\theta =\frac{\pi }{4}}}=\frac{-3}{2} $ .
$ \therefore $
Equation of tangent is $ ( y-\frac{3}{2\sqrt{2}} )=\frac{-3}{2},( x-\frac{1}{\sqrt{2}} ) $
therefore $ 3\sqrt{2}x+2\sqrt{2}y=6 $
therefore $ 3x+2y=3\sqrt{2} $ .
BETA
