A) \[2x+3y=3\sqrt{2}\]
B) \[2x-3y=3\sqrt{2}\]
C) \[3x+2y=3\sqrt{2}\]
D) \[3x-2y=3\sqrt{2}\]
Correct Answer: C
Solution :
\[{{\left. x\, \right|}_{\theta =\frac{\pi }{4}}}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}},\] \[{{\left. y\, \right|}_{\theta =\frac{\pi }{4}}}=\frac{3}{2\sqrt{2}},{{\left. \frac{dy}{dx} \right|}_{\theta =\frac{\pi }{4}}}{{\left. \frac{9\,{{\sin }^{2}}\theta \cos \theta }{-6\,{{\cos }^{2}}\theta \sin \theta } \right|}_{\theta =\frac{\pi }{4}}}=\frac{-3}{2}\]. \[\therefore \] Equation of tangent is \[\left( y-\frac{3}{2\sqrt{2}} \right)=\frac{-3}{2}\,\left( x-\frac{1}{\sqrt{2}} \right)\] Þ \[3\sqrt{2}x+2\sqrt{2}y=6\]Þ \[3x+2y=3\sqrt{2}\].
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