∴ The equation of the tangent to the parabola S ≡ y2 – 4ax = 0
at P(x1, y1) is S1 ≡ yy1 – 2a(x +
x1) = 0.
Theorem - V
The equation of the normal at P(x1, y1) on S = 0 is (y – y1) =
– (x
– x1).
Proof :
By Theorem - III, the equation of the tangent to the parabola y2 – 4ax = 0
at P(x1, y1) is S 1 ≡ yy1 – 2a(x +
x1) = 0.
∴ Slope of the tangent at P is
∴ Slope of the normal at P is – .
Hence equation of the normal at P(x1, y1) is (y – y1)
= – (x
– x1).
Theorem - VI
Equation of the tangent to the parabola y2 = 4ax at a point 't' is x – yt +
at2 = 0.
Proof :
Let P(t) be a point on the parabola y2 = 4ax then P = (at2, 2at).
We have equation of the tangent at P(x1, y1) is yy1 –
2a (x + x1) = 0, then replacing (x1, y1) by
(at2, 2at), the equation of the tangent is 2at y – 2a (x + at2)
= 0
i.e., x – yt + at2 = 0.
Theorem - VII
Equation of the normal to the parabola y2 = 4ax at a point 't' is y + xt = 2at +
at3.
Proof :
Let P(t) be a point on the parabola y2 = 4ax then P = (at2, 2at).
We have equation of the normal at P(x1, y1) is (y – y1) = –
(x –
x1) then replacing (x1, y1) by
(at2, 2at), the equation of normal is