Theorems
Theorem - I
The condition for a straight line y = mx + c to be a tangent to the parabola y2 = 4ax is cm = a
Proof :
The 'x' coordinates of the points of intersection of the line y = mx + c and the given parabola are given by the equation (3) of figure i.e.,
m2x2 + 2x (mc – 2a) + c2 = 0 ...... (1)
The given line will touch the parabola ⇔ the two points coincide.
⇔ discriminant of (1) is zero.
4(mc – 2a)2 – 4m2c2 = 0
16a (a – mc) = 0
a – mc = 0
a = mc or c =
Theorem - II
Two tangents can be drawn from an external point (x1, y1) to the parabola y2 = 4ax.
Proof :
Let P(x1, y1) be an external point to the parabola y2 = 4ax then S ≡ y12 – 4ax1 > 0 .... (1)
We have y = mx + is a tangent to the parabola y2 = 4ax for all non zero value of m.
If it passes through the point (x1, y1) then y1 = mx1 + or m2x1 – my1 + a = 0 and its discriminant y12 – 4ax1 > 0 [from (1)].
The equation being quadratic in m, has two distinct real roots say m1 and m2.
Then y = m1x + and y = m2x + are two distinct tangents through (x1, y1)
Theorem - III
The equation of the chord joining the points (x1, y1) and (x2, y2) on S = 0 is S1 + S2 = S12
Proof :
Let P(x1, y1) and Q(x2, y2) be two points on the parabola S ≡ y2 – 4ax = 0, then S11 = 0 and S22 = 0.
Consider the first degree equation S1 + S2 = S12
i.e., {yy1 – 2a(x + x1)} + {yy2 – 2a(x + x2)} = y1y2 – 2a(x1 + x2)------ (i)
i.e., 4ax – (y1 + y2)y + y1y2 = 0 which represents a straight line.
Substituting (x1, y1) in (i), we get S11 + S12 = 0 + S12 = S12.
∴ (x1, y1) satisfies the equation S1 + S2 = S12.
Similarly (x2, y2) satisfies the equation S1 + S2 = S12.
∴ S1 + S2 = S12 is a straight line passing through P(x1, y1) and Q(x2, y2).
∴ The equation of the chord PQ is S1 + S2 = S12.
Theorem - IV
The equation of the tangent at P(x1, y1) to the parabola S = 0 is S1 = 0.
Proof :
Let P(x1, y1) and Q(x2, y2) be two points on the parabola S ≡ y2 – 4ax = 0, then S11 = 0 and S22 = 0.
By Theorem - II the equation of the chord PQ is S1 + S2 = S12 .... (1)
The chord PQ becomes the tangent at P when Q approaches P (i.e., (x2, y2) approaches to (x1, y1)).
∴ The equation of the tangent at P is obtained by taking limits as (x2, y2) tends to (x1, y1) on either sides of (1).
So the equation of the tangent at P given by .
i.e., S1 + S1 = S11 [∴ S2 → S1, S12 → S11 as (x2, y2) → (x1, y1)]
∴ 2S1 = 0 ⇒ S1 = 0.
∴ 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
(y – 2at) = (x – at2)
i.e., y + xt = 2at + at3.