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.