If $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$

Question

if $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$

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My try :

$2pp' = (b^2-a^2)\sin 2\theta$

$p'^2 + pp'' = (b^2-a^2)\cos 2\theta$

Thats it ! it doesn't simplify no matter what I try. Any help ?

Answer 1

Since $$p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta$$ we have $$p=\mu (a^2 \cos^2 \theta+b^2 \sin^2 \theta)^{1/2}, \mu=\pm 1$$

$$p + p'' =\frac{\mu a^2b^2(\cos^4\theta+\sin^4 \theta+2\cos^2\theta \sin^2\theta)}{(a^2 \cos^2 \theta+b^2 \sin^2 \theta)^{3/2}}=\frac{a^2b^2}{p^3}$$

Answer 2

We need to establish $$p+p''=\frac{a^2b^2}{p^3}$$

Multiplying by $2p'(\ne0),$ $$(p+p'')2p'=a^2b^2\cdot\frac{2p'}{p^3}$$

Integrating we get, $$p^2+p'^2=-\frac{a^2b^2}{p^2}+K\text{, where }K\text{ is an arbitrary constant} $$

Multiplying by $p^2(\ne0),$ $$p^4+p'^2p^2=-a^2b^2 +p^2\cdot K\ \ \ \ (1)$$

Now, $$p^4+p'^2p^2=(p^2)^2+(pp')^2=(a^2\cos^2\theta+b^2\sin^2\theta)^2+[(b^2-a^2)\sin\theta\cos\theta]^2$$

$$=a^4\cos^2\theta(\cos^2\theta+\sin^2\theta)+b^4\sin^2\theta(\cos^2\theta+\sin^2\theta)$$

$$=a^2(p^2-b^2\sin^2\theta)+b^2(p^2-a^2\cos^2\theta)$$

$$\implies p^4+p'^2p^2=p^2(a^2+b^2)-a^2b^2$$

Comparing with $(1),K=a^2+b^2$ which definitely a constant independent of $\theta$

Answer 3

Consider values of Sin^2 (x) , Cos^2(x) and dp/dx from the given equation.

p^2 = a^2(1-Sin^2x) + b^2Sin^2x 》Sin^2x = (p^2-a^2)/ (b^2-a^2)

》p^2 = a^2 Cos^2x + b^2 (1-Cos^2x)

》Cos^2x = (b^2-p^2)/(b^2-a^2)

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now differentiate the equation with respect to x

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》p dp/dx =( b^2-a^2) Sinx. Cosx … (1)

》dp/dx = [(b^2-a^2) /p ] Sinx. Cosx

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again differentiate the equation 1 with respect to x

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》p d^2p/dx^2 + (dp/dx)^2 = (b^2-a^2)Cos^2x-Sin^2x

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now put values ofSin^2x and Cos^2x

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》(b^2-a^2)[b^2-p^2-p^2+a^2]/(b^2-a^2)

》a^2 + b^2- 2p^2

》p . (d^2p/dx^2) + (dp/dx)^2 = a^2+b^2 -2p^2

》p . (d^2p/dx^2) = a^2+b^2-(dp/dx)^2-2p^2

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put value of dp/dx

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》p d^2p / dx^2 = a^2+b^2 - [(b^2-a^2)^2 (Sinx.Cosx)^2]p^2 -2p^2

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now put p^2 = a^2Cos^2x+b^2Sin^2x

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》p d^2p / dx^2 = p^2 (a^2+b^2) - [{b^4+a^4-2.a^2.b^2}{Sin^2(x).Cos^2 (x)}] / p^2 - 2 p^2

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p^2 = a^2Cos^2x +b^2Sin^2x

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》p d^2p / dx^2 = (a^2Cos^2x + b^2Sin^2x)(a^2+b^2)-b^4.Sin^2x.Cos^2x - a^4.Sin^2x.Cos^2x + 2a^2.b^2.Sin^2x.Cos^2x/p^2

-2p^2

》 p d^2p / dx^2 = [a^4.Cos^2x + a^2.b^2.Cos^2x + b^2.a^2Sin^2x + b^4.Sin^2x -a^4.Cos^2x. (1-Sin^2x) -b^4.Sin^2x (1-Sin^2x) + 2a^2.b^2Sin^2x.Cos^2x ] p^2 - 2p^2

》 p d^2p / dx^2 = [a^4.Cos^2x + b^4.Sin^2x + a^2.b^2(Sin^2x+Cos^2x) - b^4.Sin^2x - a^4.Cos^2x + a^4.Cos^4x + b^4.Sin^4x + 2a^2.b^2.Sin^2x.Cos^2x] p^2 -2p^2

》p d^2p / dx^2 = [a^2.b^2 + (b^2.Sin^2x)^2+ (a^2.Cos^2x)^2 + 2a^2.b^2.Sin^2x.Cos^2x] / p^2 - 2p^2

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(a+b)^2 = a^2 + b^2 + 2 a^2.b^2

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》 p d^2p / dx^2 = [a^2.b^2 + (a^2.Cos^2x +b^2.Sin^2x)^2 ] / p^2 - 2p^2

》 p d^2p / dx^2 = [a^2.b^2 + (p^2)^2 ] / p^2 - 2p^2

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separating in per terms

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》 p d^2p / dx^2 = a^2.b^2/p^2 + p^4/p^2 -2p^2

》p d^2p/dx^2 = a^2.b^2/p^2 - p^2

》 d^2p / dx^2 = a^2.b^2/p.p^2 - p^2/p

》 d^2p / dx^2 = a^2.b^2/p^3 - p

| d2p / dx^2 + p = a^2b^2/p^3 |

full explained ...hope it help.