Answer to Question #226667 in Analytic Geometry for Anuj

Question #226667

The coordinates of the ends of a focal chord of the parabola y2= 4ax are (x1, y1) and (x2, y2). Show that x1x2 = a2 and y1y2 = −4a2.


1
Expert's answer
2021-08-19T11:04:41-0400

Say point A (x1, y1) = (ap12,2ap1)(ap_1^2, 2ap_1) , and

point B (x2, y3) = (ap22,2ap2)(ap_2^2, 2ap_2) coordinates of the ends of a focal chord of the

parabola y2=4axy^2 = 4ax


Then O, A and B are collinear, where O is the vertex.


Therefore, Slope of OA = Slope of OB


Or, 2ap1(ap12a)=2ap2(ap22a)\frac{2ap_1}{(ap_1^2-a)} = \frac{2ap_2}{(ap_2^2-a)}


On solving,, we get


p1p22p1=p12p2p2p_1p_2^2 - p_1 = p_1^2p_2 - p_2


p1p2(p2p1)=(p2p1)p_1 p_2(p_2-p_1) = -(p_2-p_1)


Or

p1p2=1p_1 p_2 = -1


Now

x1x2=ap12.ap22x_1 x_2 = ap_1^2 . ap_2^2

Or

x1x2=a2(1)=a2x_1 x_2 = a^2 (1) = a^2

Therefore,

x1x2=a2(Proved)x_1 x_2 = a^2 (Proved)


Now

y1y2=2ap12ap2=4a2p1p2=4a2y_1 y_2 = 2ap_1 2a p_2 = 4a^2p_1p_2 = -4a^2

Therefore,

y1y2=4a2(Proved)y_1 y_2 = -4a^2 (Proved)


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