A) \[y=x\]
B) \[y=x+2\]
C) \[y=-x+2\]
D) \[y=-x+3\]
Correct Answer: A
Solution :
Given curves is \[{{x}^{3}}+{{y}^{3}}=2xy\] ?.(i) If a line is a tangent of given curve, then it will touch at only one point. For \[y=x\] to be a tangent, \[{{x}^{3}}+{{x}^{3}}=2x\times x\Rightarrow 2{{x}^{3}}=2{{x}^{2}}\] \[\Rightarrow \] \[x=1\] On putting the value of x in given curve (i), we get \[{{(1)}^{3}}+{{y}^{3}}=2\times 1\times y\] \[\Rightarrow \] \[1+{{y}^{3}}=2y\] \[\Rightarrow \] \[{{y}^{3}}-2y+1=0\] \[\Rightarrow \] \[{{y}^{3}}-{{y}^{2}}+{{y}^{2}}-2y+1=0\] \[\Rightarrow \] \[{{y}^{2}}(y-1)+{{y}^{2}}-y+y-2y+1=0\] \[\Rightarrow \] \[{{y}^{2}}(y-1)+y(y-1)-y+1=0\] \[\Rightarrow \] \[{{y}^{2}}(y-1)+y(y-1)-1(y-1)=0\] \[\Rightarrow \] \[(y-1)({{y}^{2}}+y-1)=0\] \[\Rightarrow \] \[y-1=0\] or \[{{y}^{2}}+y-1=0\] \[\Rightarrow \] \[y=1\] or \[y=\frac{-1\pm \sqrt{1-4\times 1\times (-1)}}{2}\] \[\Rightarrow \] \[y=1\] or \[y=\frac{-1\pm \sqrt{5}}{2}\][not integral value] So, \[(x,\,\,\,y)\,\,=\,\,\,(1,\,\,1)\] Then, \[y=x\] will be the tangent to the given curve.
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