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Analytical Geometry

Blazing goIITian

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4 Apr 2009 20:57:12 IST

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Through the vetex A of the parabola y^2=4ax two chords AP and AQ are drawn, and the circles on AP an

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Through the vetex A of the parabola y^2=4ax two chords AP and AQ are drawn, and the circles on AP and AQ as diameters intersect in R. Prove that, if theta1, theta 2, and phi be the angles made with the axis by the tangents at P and Q and by AR, then cot theta1+cot theta2 +2 tan phi=0.

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Prakhar Banga

Blazing goIITian

Joined: 20 Dec 2008

Posts: 599

4 Apr 2009 21:17:27 IST

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Let P be (at^2, 2at) and Q be (am^2, 2am).

As the equation of a circle with (x1, y1) and (x2, y2) as diameters is given by (x-x1)(x-x2)+(y-y1)(y-y2) = 0, so the two circles are x(x-at^2) + y(y-2at) = 0 and x(x-am^2) + y(y-2am) = 0.

Let they intersect at R (h, k).

h(h-at^2) + k(k-2at) = 0

h(h-am^2) + k(k-2am) = 0

Subtracting the two

ah(t^2-m^2) + 2ka(t-m) = 0.

t is not equal to m.

So h(t+m) + 2k = 0

h = -2k/(t+m).......1

k/h = -(t+m)/2

This is the slope of AR = tan phi(As A is the Origin)

The slopes of tangents at (at^2, 2at) and (am^2, 2am) are 1/t and 1/m.

So the value of cot theta1 and cot theta2 are t and m

=> cot theta1 + cot theta2 + 2 tan phi = t + m + 2(-t-m) /2 = 0

Hence proved.

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