jmc

Question

The points P = (x_1,y_1) and Q = (x_2,y_2) are the intersections of the parabola y^2 = 4ax, where a > 0, and a line passing through the focus of the parabola. Then the distance PQ can be expressed in the form c_1 x_1 + c_2 x_2 + c_3 a, where c_1, c_2, and c_3 are constants. Compute c_1 + c_2 + c_3.

Accepted Answer

The focus of the parabola y^2 = 4ax is F = (a,0), and the directrix is x = -a. Then PQ = PF + QF.FIGURE Since P lies on the parabola, PF is equal to the distance from P to the directrix, which is x_1 + a. Similarly, QF is equal to the distance from Q to the directrix, which is x_2 + a. Therefore, PQ = x_1 + x_2 + 2a.Hence, c_1 + c_2 + c_3 = 1 + 1 + 2 = 4.