Question

If p1 and p be the radius of curvature at the end of a focal chord of the parabola y2 = 4ax, then show that p1(-2/3)+p2(-2/3) = (2a)(-2/3)

Answer 1

Step-by-step explanation:

To solve this problem, let us take the parametric equation of the parabola y² = 4ax,

x = at², y = 2at

Then dx/dt = 2at, d²x/dt² = 2a

and dy/dt = 2a, d²y/dt² = 0

Therefore radius of curvature,

p = {(dx/dt)² + (dy/dt)²}^(3/2) / (dx/dt d²y/dt² - dy/dt d²x/dt²)

= {4a²t² + 4a²}^(3/2) / (0 - 4a²)

= - {4a² (t² + 1)}^(3/2) / 4a²

= {4a² (t² + 1)}^(3/2) / 4a² [ in magnitude ]

= 8a³ (1 + t²)^(3/2) / 4a²

= 2a (1 + t²)^(3/2)

Let 't₁' and 't₂' be the ends of a focal chord of the parabola.

Then t₁ t₂ = - 1

Thus p₁ = 2a (1 + t₁²)^(3/2)

and p₂ = 2a (1 + t₂²)^(3/2)

∴ p₁^(-2/3) + p₂^(- 2/3)

= (2a)^(- 2/3) {1/(1 + t₁²) + 1/(1 + t₂²)}

= (2a)^(- 2/3)

i.e., p₁^(-2/3) + p₂^(- 2/3) = (2a)^(- 2/3)

Hence proved.

Calculation:

1/(1 + t₁²) + 1/(1 + t₂²)

= 1/(1 + t₁²) + 1/{1 + (- 1/t₁)²} [ ∵ t₁ t₂ = - 1 ]

= 1/(1 + t₁²) + t₁²/(1 + t₁²)

= (1 + t₁²)/(1 + t₁²)

= 1

Radius of curvature ( ρ ) formulae:

1. For Cartesian equation y = f(x)

ρ = (1 + y₁²)^(3/2) / y₂

where y₂ ≠ 0

2. For parametric equation x = φ(t), y = ψ(t)

ρ = (x'² + y'²)^(3/2) / (x' y" - y' x")

where x' y" - y' x" ≠ 0

3. For polar equation r = f(θ)

ρ = (r² + r₁²)^(3/2) / (r² + 2 r₁² - r r₂)

where r² + 2 r₁² - r r₂ ≠ 0

4. For pedal equation p = f(r)

ρ = r dr/dp

5. For tangential polar equation p = f(ψ)

ρ = p + d²p/dψ²