The slope of normal at any point (x, y), x > 0, y > 0 on the curve y = y(x) is given by x²/(xy - x²y² - 1). If the curve passes through the point (1, 1), then e . y(e) is equal to

Solution

Maths sir · 4 steps

Turn the normal's slope into the tangent's slope, then put v = xy and the equation separates neatly.

1. Slope of tangent. The tangent slope is -1 divided by the normal slope. So dy/dx = -(xy - x²y² - 1)/x² = (1 + x²y² - xy)/x².
2. Put v = xy. The terms xy and x²y² suggest v = xy. Then dv/dx = y + xdy/dx = v/x + (1 + v² - v)/x = (1 + v²)/x, and the v terms cancel.

   ∫ dv/(1+v²) = ∫ dx/x ⇒ tan⁻¹(xy) = ln x + C

3. Find the constant. The curve passes through (1, 1), so tan⁻¹ 1 = ln 1 + C. That gives C = π/4.
4. So. At x = e, tan⁻¹(e y(e)) = 1 + π/4. So e y(e) = tan(π/4 + 1) = (1 + tan 1)/(1 - tan 1), which is option (d).

The idea to keep. When xy keeps appearing, try v = xy, because dv/dx = y + xdy/dx. And always convert a normal's slope to the tangent's slope with -1/m before you set up dy/dx.

Answer from the NTA final answer key. Solution by PrepWiser.