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CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 8 of 50
Marks: +1, -0
If (x2+y2)2=xy(x^2+y^2)^2 = xy, then   dydx\; \frac{dy}{dx} is
Solution:     👈: Video Solution
Explanation: Given, (x2+y2)2=xy(x^2+y^2)^2 = xy
⇒    x4+2x2y2+y4−xy=0\Rightarrow\;\; x^4+2x^2y^2+y^4-xy=0
Differentiating w.r.t. xx, we get
4x3+2[2xy2+x2⋅2y  dydx]+4y3  dydx−[y+x  dydx]=04x^3+2\left[2xy^2+x^2\cdot2y\;\frac{dy}{dx}\right]+4y^3\;\frac{dy}{dx}-\left[y+x\;\frac{dy}{dx}\right]=0
  dydx[4x2y+4y3−x]+[4x3+4xy2−y]=0\;\frac{dy}{dx}\left[4x^2y+4y^3-x\right]+\left[4x^3+4xy^2-y\right]=0
  dydx=  −[4x3+4xy2−y][4x2y+4y3−x]\;\frac{dy}{dx}=\;\frac{-\left[4x^3+4xy^2-y\right]}{\left[4x^2y+4y^3-x\right]}
   or       dydx=  y−4x(x2+y2)4y(x2+y2)−x\;\text{ or }\;\;\;\frac{dy}{dx}=\;\frac{y-4x(x^2+y^2)}{4y(x^2+y^2)-x}
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