A) \[(2,2)\]
B) \[(-1,-2)\]
C) \[\left( \frac{9}{4},\frac{3}{8} \right)\]
D) None of these
Correct Answer: C
Solution :
Given that, \[{{y}^{2}}=x{{(2-x)}^{2}}\] On differentiating w.r.t.x, we get \[2y\frac{dy}{dx}={{(2-x)}^{2}}+2x(2-x)(-1)\] At point \[(1,\,1)\] \[2(1)\frac{dy}{dx}={{1}^{2}}-2\] \[\frac{dy}{dx}=-\frac{1}{2}\] \[\therefore \] Equation of tangent at \[(1,\,\,1)\] is \[y-1=-\frac{1}{2}(x-1)\] \[\Rightarrow \] \[y=-\frac{x}{2}+\frac{3}{2}\] On solving Eqs (i) and (ii), we get \[x=\frac{9}{4},\,\,y=\frac{3}{8}\]
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