Let the extremities of the normal chord be P and Q and the tangents at P and Q to the parabola say y 2 = 4ax meet in T. Let the co-ordinates of T be (x 1 , y 1 ).

PQ will be the chord of contact for T with respect to the parabola so area of triangle TPQ will be

= ...... (1)

The equation to the chord of contact of T will be

yy 1 = 2a (x+x 1 )

or y = ...... (2)

Equation to any normal to y 2 = 4ax is

y = mx – 2am –am 3 .......(3)

So (2) and (3) must be identical. As the coefficients of y are equal, others must also be equal, so

m = and – 2am – am 3 =

when y 1 = and x 1 = (–2a – am 2 ).

If the inclination of the chord of contact, i.e., the normal is θ ; then

m = tan θ

So y 1 = =2a cot θ and x 1 = (–2a –a tan 2 θ ).

Substituting in (1), we get

Area of the triangle =

= =

=

= 4a 2 sec 3 θ . cosec 3 θ . Proved

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