The slope of tangent at any point (x, y) on a curve y=y(x) is (x² + y²)/2xy,x > 0. If y(2) = 0, then a value of y(8) is :

Solution

Maths sir · 4 steps

Every term has degree 2, so put y = vx and the equation separates neatly into v and x.

1. Homogeneous substitution. The right side (x² + y²)/2xy is homogeneous, so let y = vx, which gives dy/dx = v + xdv/dx. Then v + xdv/dx = (1 + v²)/2v, so xdv/dx = (1 - v²)/2v.

   ∫ 2v/(1 - v²) dv = ∫ dx/x ⇒ -ln|1 - v²| = ln x + C

2. Return to y. This gives x(1 - v²) = k for a constant k. Put v = y/x and multiply by x to get x² - y² = kx.
3. Use the point. At x = 2, y = 0, so 4 = 2k and k = 2. The curve is y² = x² - 2x.
4. So. At x = 8, y² = 64 - 16 = 48, so y = ± 4√3. The value among the options is 4√3, option (c).

The idea to keep. When every term in dy/dx has the same degree in x and y, substitute y = vx. The equation then separates, and you fix the constant with the given point.

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