The curve amongst the family of curves represented by the differential equation, , which passes through , is
- A.a circle with centre on the Y-axis
- B.a circle with centre on the X-axis
- C.an ellipse with major axis along the Y-axis
- D.a hyperbola with transverse axis along the X-axis
Show answer and solution
Correct option: B
Solution
Rewrite the ODE as dy/dx = (y² - x²)/(2xy), which is homogeneous. Substitute y = vx, so dy/dx = v + x(dv/dx).
Then: v + x(dv/dx) = (v²x² - x²)/(2x·vx) = (v² - 1)/(2v)
x(dv/dx) = (v² - 1)/(2v) - v = (v² - 1 - 2v²)/(2v) = -(v² + 1)/(2v)
Separate: 2v/(v² + 1) dv = -dx/x
Integrate: ln(v² + 1) = -ln x + C, so (v² + 1)x = k.
Substitute back v = y/x: (y²/x² + 1)x = k → (y² + x²)/x = k → x² + y² = kx, i.e., x² - kx + y² = 0.
This is x² + y² = kx, a circle with centre at (k/2, 0) — on the X-axis.
Applying the condition (1,1): 1 + 1 = k·1, so k = 2. The curve is x² + y² = 2x, a circle centred at (1, 0) on the X-axis.
Then: v + x(dv/dx) = (v²x² - x²)/(2x·vx) = (v² - 1)/(2v)
x(dv/dx) = (v² - 1)/(2v) - v = (v² - 1 - 2v²)/(2v) = -(v² + 1)/(2v)
Separate: 2v/(v² + 1) dv = -dx/x
Integrate: ln(v² + 1) = -ln x + C, so (v² + 1)x = k.
Substitute back v = y/x: (y²/x² + 1)x = k → (y² + x²)/x = k → x² + y² = kx, i.e., x² - kx + y² = 0.
This is x² + y² = kx, a circle with centre at (k/2, 0) — on the X-axis.
Applying the condition (1,1): 1 + 1 = k·1, so k = 2. The curve is x² + y² = 2x, a circle centred at (1, 0) on the X-axis.
[JEE Main 2019 · 10 Jan · Shift II]