A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with :

Solution

Maths sir · 5 steps

Turn the midpoint condition into a differential equation, solve it, and the parabola falls out.

1. Find Q. Let P = (x, y) and slope y' = dy/dx. The tangent Y - y = y'(X - x) meets the x-axis where Y = 0, so Q = (x - y/y', 0).
2. Use the bisection. The y-axis bisects PQ, so the midpoint of PQ has x-coordinate 0. That means x + x - y/y' = 0, which gives dy/dx = y/2x.

   ∫ (2 dy)/y = ∫ dx/x ⇒ y² = kx

3. Fix the constant. The curve passes through (3, 3), so 9 = 3k and k = 3. The curve is y² = 3x.
4. Read off features. Compare with y² = 4ax: 4a = 3, so a = 3/4. The latus rectum is 4a = 3 and the focus is (3/4, 0).
5. So. The parabola y² = 3x has latus rectum of length 3, which is option (a).

The idea to keep. When a geometric condition on the tangent defines a curve, write the tangent's intercept in terms of x, y and y', form the differential equation, then use the given point to fix the constant.

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