Question

find the angle between radius vector and tangent
rsec^2 (theta/2)=2a​

Answer 1

The angle between the radius vector and tangent is ∅ = $\frac{\pi }{2}$ + θ

Given that;

r$sec^{2}$(θ/2) = 2a

To find;

The angle between the radius vector and tangent.

Solution;

We have r$sec^{2}$(θ/2) = 2a,

Taking log on both sides we get,

log r  + log$sec^{2}$(θ/2) = log 2a

log r + 2logsec(θ/2) = log 2a

Differentiating w.r.t to θ,

$\frac{1}{r}$ dr/dθ + 2 . 1/sec(θ/2) . sec(θ/2) . tan(θ/2) . $\frac{1}{2}$ = 0

$\frac{1}{r}$ dr/dθ + tan(θ/2) = 0

Since, we know that cot($\frac{\pi }{2}$ + θ) = - tan θ and cot ∅ = $\frac{1}{r}$ dr/dθ

cot ∅ = cot( $\frac{\pi }{2}$ + θ/2)

∅ = $\frac{\pi }{2}$ + θ/2

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