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JEE Mains · Maths · STD 12 - 9. differential equations

The slope of normal at any point \((x, y), x > 0, y > 0\) on the curve \(y=y(x)\)  is given by \(\frac{x^{2}}{x y-x^{2} y^{2}-1}\) If the curve passes through the point \((1,1)\), then e. \(y(e)\) is equal to

  1. A \(\frac{1-\tan (1)}{1+\tan (1)}\)
  2. B \(\tan (1)\)
  3. C \(1\)
  4. D \(\frac{1+\tan (1)}{1-\tan (1)}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{1+\tan (1)}{1-\tan (1)}\)

Step-by-step Solution

Detailed explanation

Slope of normal \(=\frac{-d x}{d y}=\frac{x^{2}}{x y-x^{2} y^{2}-1}\) \(x^{2} y^{2} d x+d x-x y d x=x^{2} d y\) \(x^{2} y^{2} d x+d x=x^{2} d y+x y d x\) \(x^{2} y^{2} d x+d x=x(x d y+y d x)\) \(x^{2} y^{2} d x+d x=x d(x y)\) \(\frac{d x}{x}=\frac{d(x y)}{1+x^{2} y^{2}}\)…
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