Let a conic C pass through the point (4,-2) and P(x, y), x ≥ 3, be any point on C. Let the slope of the line touching the conic C only at a single point P be half the slope of the line joining the points P and (3,-5). If the focal distance of the point (7,1) on C is d, then 12 d equals ________.

Solution

Maths sir · 5 steps

Turn the slope condition into a differential equation, solve it, and a shifted parabola appears.

1. Set up the equation. The tangent at P(x, y) has slope dy/dx. The line from P to (3,-5) has slope (y+5)/(x-3). The tangent slope is half of this.

   dy/dx = 1/2·(y+5)/(x-3) ⇒ (2 dy)/(y+5) = dx/(x-3)

2. Integrate. This gives 2 ln|y+5| = ln|x-3| + const, that is, (y+5)² = c(x-3). The point (4,-2) gives 9 = c × 1, so c = 9 and C is (y+5)² = 9(x-3).
3. Read the parabola. With X = x-3 and Y = y+5 this is Y² = 4aX with 4a = 9, so a = 9/4. The vertex is (3,-5). The point (7,1) lies on C, since 6² = 36 = 9 × 4.
4. Focal distance. For Y² = 4aX the focal distance of a point is X + a. At (7,1), X = 7 - 3 = 4, so d = 4 + 9/4 = 25/4.
5. So. 12d = 12 × 25/4 = 75.

The common slip: 111. Using the original x = 7 in x + a instead of the shifted X = 4 gives d = 37/4 and 12d = 111, not 75.

The idea to keep. A slope condition at every point is a differential equation, so separate the variables and solve it. For a shifted parabola, measure the focal distance X + a from the vertex, not from the origin.

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