Question:

The solution of \[ (1+y^2)\,dx-xy\,dy=0,\quad y(1)=0 \] represents a conic. Its eccentricity is

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For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] eccentricity is \[ e=\sqrt{1+\frac{b^2}{a^2}}. \]
Updated On: Jun 25, 2026
  • \(2\)
  • \(\dfrac{1}{e}\)
  • \(1\)
  • \(\sqrt{2}\)
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The Correct Option is D

Solution and Explanation

Step 1: Rearrange the differential equation.
Given: \[ (1+y^2)\,dx-xy\,dy=0 \] Rearranging, \[ (1+y^2)\,dx=xy\,dy \] Therefore, \[ \frac{dx}{x}=\frac{y}{1+y^2}\,dy \]

Step 2: Integrate both sides.
Integrating, \[ \int \frac{1}{x}\,dx = \int \frac{y}{1+y^2}\,dy \] \[ \log x = \frac12\log(1+y^2)+C \] Multiply by \(2\): \[ 2\log x=\log(1+y^2)+C \] Using logarithm properties: \[ \log x^2=\log(1+y^2)+C \] Hence, \[ x^2=C(1+y^2) \]

Step 3: Use the initial condition.
Given: \[ y(1)=0 \] Substitute: \[ 1=C(1+0) \] \[ C=1 \] Therefore, \[ x^2=1+y^2 \] or \[ x^2-y^2=1 \]

Step 4: Identify the conic.
The equation \[ x^2-y^2=1 \] represents a hyperbola.
Comparing with \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] we get \[ a^2=1,\quad b^2=1 \] Eccentricity of hyperbola is \[ e=\sqrt{1+\frac{b^2}{a^2}} \] So, \[ e=\sqrt{1+\frac11} \] \[ e=\sqrt{2} \]

Step 5: Final conclusion.
Therefore, the eccentricity is \[ \boxed{\sqrt{2}} \]
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