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 :
Given curves are \[x=2{{\cos }^{3}}\theta \] and \[y=3{{\sin }^{3}}\theta \] \[\therefore \] \[\frac{dx}{d\theta }=-6{{\cos }^{2}}\theta \,\sin \theta ,\frac{dy}{d\theta }=9{{\sin }^{2}}\theta \cos \theta \] \[\therefore \]\[\frac{dy}{dx}=-\frac{9{{\sin }^{2}}\theta \,\cos \theta }{6\,{{\cos }^{2}}\,\theta \,\sin \theta }=-\frac{3}{2}\tan \theta \] At \[\theta =\frac{\pi }{4},\frac{dy}{dx}=-\frac{3}{2}\tan \frac{\pi }{4}=-\frac{3}{2}\] Also, at \[\alpha =\frac{\pi }{4},x=2{{\cos }^{3}}\frac{\pi }{4}=2{{\left( \frac{1}{\sqrt{2}} \right)}^{3}}=\frac{1}{\sqrt{2}}\] and \[y=3{{\sin }^{3}}\theta =3{{\left( \frac{1}{\sqrt{2}} \right)}^{3}}=\frac{3}{2\sqrt{2}}=\frac{3\sqrt{2}}{4}\] Equation of tangent at \[\left( \frac{1}{\sqrt{2}},\frac{3\sqrt{2}}{4} \right)\] is \[\left( y-\frac{3\sqrt{2}}{4} \right)=-\frac{3}{2}\left( x-\frac{1}{\sqrt{2}} \right)\] \[\Rightarrow \] \[\frac{(4y-3\sqrt{2})}{4}=-\frac{3}{2}\left( \frac{2x-\sqrt{2}}{2} \right)\] \[\Rightarrow \] \[4y-3\sqrt{2}=-6x+3\sqrt{2}\] \[\Rightarrow \] \[4y+6x=3\sqrt{2}+3\sqrt{2}\] \[\Rightarrow \] \[4y+6x=3\sqrt{2}+3\sqrt{2}\]
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