Question

If p² = a²cos²θ + b²sin²θ, then prove that p + d²p/dθ² = a²b²/p³. 

Answer 1

We have p² = a²cos²θ + b²sin²θ

⇒ 2p² = a²(2cos²θ) + b²(2sin²θ)

⇒ 2p² = a²(1 + cos²θ) + b²(1 – cos²θ)

⇒ 2p² = (a² + b²) + (a² – b²)cos²θ

Differentiate w.r.t q, we get,

= (a²cos²θ + b²sin²θ)(b²cos²θ + a²sin²θ) – (b² – a²)sin²θcos²θ

= a²b²(cos⁴θ + sin⁴θ + 2sin²θcos²θ)

= a² b²(sin²θ + cos²θ)² = a²b²

Thus, p + d²p/dθ² = a²p²/p³