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
- A \(\frac{1-\tan (1)}{1+\tan (1)}\)
- B \(\tan (1)\)
- C \(1\)
- D \(\frac{1+\tan (1)}{1-\tan (1)}\)
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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