If P_1 P_2 and P_3 P_4 are two focal chords o | vedclass

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(A) Let the coordinates of the points be $P_i(at_i^2, 2at_i)$ for $i = 1, 2, 3, 4$. Since $P_1 P_2$ and $P_3 P_4$ are focal chords, we have $t_1 t_2 =

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directrix of the parabola

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(A) Let the coordinates of the points be $P_i(at_i^2, 2at_i)$ for $i = 1, 2, 3, 4$. Since $P_1 P_2$ and $P_3 P_4$ are focal chords, we have $t_1 t_2 = -1$ and $t_3 t_4 = -1$. The equation of the chord joining $P_i$ and $P_j$ is $(t_i + t_j)y = 2x + 2at_i t_j$. For $P_1 P_3$, the equation is $(t_1 + t_3)y = 2x + 2at_1 t_3$ ... $(1)$. For $P_2 P_4$, the equation is $(t_2 + t_4)y = 2x + 2at_2 t_4$ ... $(2)$. Since $t_2 = -1/t_1$ and $t_4 = -1/t_3$, substituting these into $(2)$ gives $( -1/t_1 - 1/t_3 )y = 2x + 2a(-1/t_1)(-1/t_3)$, which simplifies to $-(t_1 + t_3)y = 2xt_1 t_3 + 2a$. Solving the system of equations $(1)$ and $(2)$ shows that the intersection point lies on the line $x = -a$, which is the directrix of the parabola.

If $P_1 P_2$ and $P_3 P_4$ are two focal chords of the parabola $y^2 = 4ax$, then the chords $P_1 P_3$ and $P_2 P_4$ intersect on the

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