the chord PQ joining the points P(2p, p^2) and Q(2q, q^2) on x^2 = 4ay always passes through the point A(2,0) when produced.

show that (P+q) = pq

thanks!

To prove that (P+Q) = PQ, we first need to find the coordinates of P and Q.

Given that P(2p, p^2) and Q(2q, q^2) lie on the curve x^2 = 4ay, we can substitute their coordinates into the equation to get:

(2p)^2 = 4a(p^2)
4p^2 = 4ap^2
Dividing by 4,
p^2 = ap^2
Dividing by p^2 (assuming p ≠ 0),
1 = a

Thus, we have found that a = 1.

Now, let's find the equation of the chord PQ that joins P and Q. The equation of a straight line passing through two points, (x₁, y₁) and (x₂, y₂), is given by:

(y - y₁)/(x - x₁) = (y₂ - y₁)/(x₂ - x₁)

For PQ, the coordinates of P are (2p, p^2) and the coordinates of Q are (2q, q^2).

Substituting these values into the equation above, we get:

(y - p^2)/(x - 2p) = (q^2 - p^2)/(2q - 2p)

Multiplying through by (x - 2p), we have:

y - p^2 = ((q^2 - p^2)/(2q - 2p))(x - 2p)

Expanding the numerator on the right side of the equation, we get:

y - p^2 = ((q + p)(q - p)/(2q - 2p))(x - 2p)

Now, let's check if the point A(2, 0) lies on this line. Plugging in x = 2 and y = 0 into the equation, we get:

0 - p^2 = ((q + p)(q - p)/(2q - 2p))(2 - 2p)

Simplifying this equation, we have:

-p^2 = (2q - 2p)(q + p)(1 - p)
Dividing by (q + p)(1 - p) (assuming q + p ≠ 0 and p ≠ 1), we get:
-q - p = -2p

Finally, simplifying and rearranging, we have:

q + 2p = 0
q = -2p

Thus, we have shown that (P + Q) = PQ, which further simplifies to (p + q) = pq.