Let a curve y=f(x), x ∈(0, ∞) pass through the points P(1, 3/2) and Q(a, 1/2). If the tangent at any point R(b, f(b)) to the given curve cuts the y-axis at the point S(0, c) such that b c=3, then (P Q)² is equal to __________.

Solution

Maths sir · 5 steps

Turn the condition bc = 3 into a differential equation, solve it, and the curve gives you Q.

1. Find the intercept. The tangent at R(b, f(b)) is y - f(b) = f'(b)(x - b). Put x = 0 to get c = f(b) - b f'(b).
2. Build the equation. Since bc = 3 for every point b, replace b by x and f(b) by y. This gives x(y - xy') = 3, which rearranges to a linear differential equation.

   dy/dx - y/x = -3/x² ⇒ d/dx(y/x) = -3/x³

3. Solve and fix C. The integrating factor is 1/x. Integrating gives y/x = 3/2x² + C, so y = 3/2x + Cx. The curve passes through P(1, 3/2), so 3/2 + C = 3/2, which gives C = 0 and y = 3/2x.
4. Locate Q. Put y = 1/2: 3/2a = 1/2, so a = 3 and Q = (3, 1/2).
5. So. (PQ)² = (3 - 1)² + (1/2 - 3/2)² = 4 + 1 = 5.

The idea to keep. When a tangent property holds at every point, write the intercept in terms of x, y and y'. That turns the property into a differential equation for the curve.

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