## Parabola – Tangent and Chord joining Two Points

The equation of the chord joining two points A (x₁, y₁), B (x₂, y₂) on the parabola S = 0 is S₁ + S₂ = S₁₂.

**Proof**: Let S = y² – 4ax = 0 be the given parabola

Since A and B are two points on the parabola

S₁₁ = 0, S₂₂ = 0

Consider the equation S₁ + S₂ = S₁₂ … (1)

yy₁ – 2a (x + x₁) + yy₂ – 2a (x + x₂) = y₁y₂ – 2a (x₁ + x₂)

it is a first degree equation in x and y and hence it represents a straight line

substituting A in equation (1), we get S₁₁ + S₁₂ = S₁₂ which is true, since S₁₁ = 0

∴ A lies on the line (1)

Substituting B in equation (1), we get S₁₂ + S₂₂ = S₁₂ which is true, since S₂₂ = 0

∴ B lies on the line (1)

Thus, the equation of AB is S₁ + S₂ = S₁₂

**Tangent: **The equation of the tangent to the parabola S = 0 at P (x₁, y₁) is S₁ = 0

**Proof: **The equation of the tangent to the parabola S = 0 at P (x₁, y₁) is S₁ = 0

Let S = y² – 4ax be the given parabola

Differentiation with respect the x

We get \(2y\frac{dy}{dx}-4a=0\).

dy/dx = 2a/y

the slop of the tangent at P is \({{\left( \frac{dy}{dx} \right)}_{p}}=\frac{2a}{{{y}_{1}}}\).

the equation of the tangent at P is \(y-{{y}_{1}}=\frac{2a}{{{y}_{1}}}(x-{{x}_{1}})\).

yy₁ – y₁² = 2ax – 2ax₁

yy₁ – 2ax = y₁² – 2ax₁

yy₁ – 2a(x + x₁) = y₁² – 4ax₁

S₁ = S₁₁

∴ S₁ = 0.