Let P, Q and R be the points at which the tangents are drawn and let their coordinates be

(at 12 , 2at 1 ), (at 22 , 2at 2 ), and (at 32 , 2at 3 ).

 tangents at Q and R intersect in the point

{at 2 t 3 , a(t 2 + t 3 )}.

Similarly, the other pairs of tangents meet at the points

{at 3 t 1 , a(t 3 + t 1 )} and {at 1 t 2 ,a (t 1 + t 2 )}.

Let the equation to the circle be

x 2 + y 2 + 2gx + 2fy + c = 0 ...... (1).

Since it passes through the above three points, we have

a 2 t 22 t 32 +a 2 (t 2 + t 3 ) 2 + 2gat 2 t 3 + 2fa (t 2 + t 3 ) + c = 0 ...... (2),

a 2 t 32 t 12 +a 2 (t 3 + t 1 ) 2 + 2gat 3 t 1 + 2fa (t 3 + t 1 ) + c = 0 ...... (3),

and a 2 t 12 t 22 +a 2 (t 1 + t 2 ) 2 + 2gat 1 t 2 + 2fa (t 1 + t 2 ) + c = 0 ...... (4),

Subtracting (3) from (2) and dividing by a (t 2 – t 1 ), we have

a{t 32 (t 1 + t 2 ) + t 1 + t 2 + 2t 3 }+ 2gt 3 + 2f = 0.

Similarly, from (3) and (4), we have

a{t 12 (t 2 + t 3 ) + t 2 + t 3 + 2t 1 } + 2gt 1 + 2f = 0.

From these two equations we have

2g = – a (1+t 2 t 3 + t 3 t 1 + t 1 t 2 ) and 2f = – a(t 1 + t 2 + t 3 – t 1 t 2 t 3 )

Substituting these values in (2), we obtain

c = a 2 (t 2 t 3 + t 3 t 1 + t 1 t 2 ).

The equation to the circle is therefore

x 2 + y 2 – ax (1 + t 2 t 3 + t 3 t 1 + t 1 t 2 ) – ay (t 1 + t 2 + t 3 – t 1 t 2 t 3 ) + a 2 (t 2 t 3 + t 3 t 1 + t 1 t 2 ) = 0.

Which clearly passes through the focus (a, 0).

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