Focal Chord:
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Any chord to the parabola y2 = 4ax which passes through focus is called as focal chord of parabola y2 = 4ax.
Let y2 = 4ax be the equation of parabola and (at2, 2at) a point P on it. Assume that the coordinates of the other extremity Q of the focal chord through P are (at12, 2at1).
Then, PS and SQ, where S is focus (a, 0) have the same slopes.
 tt12 - t = t1 t2 - t1 => (tt1 + 1)(t1 - t) = 0.
Hence t1 = -1/t, that is the point Q is (a/t2, -2a/t).
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That is the extremities of a focal chord of parabola y2 = 4ax may be taken as points t and -1/t.
Example: Prove that circle with any focal chord of parabola y2 = 4ax as its diameter always touches the directrix of it.
Solution: Let AB be the focal chord. If A is (at2, 2at), then B is 
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Equation of circle having AB as diameter is
For x = -a, this gives
- 2ay(t - 1/t) = 0
=> [y - a(t - 1/t)]2 = 0, which has equal roots.
=> x + a = 0 is a tangent to circle having diameter AB.
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