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Q.
Which of the following is a tangent to the curve given by $ {{x}^{3}}+{{y}^{3}}=2xy? $
J & K CETJ & K CET 2015Application of Derivatives
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.