Find the value of ## (ax){^\frac{2}{3}} + (by){^\frac{2}{3}}##

[sahilmm15]

Homework Statement

Given, if ${\frac{ax}{cos\theta}} + {\frac{by}{sin\theta}} = a^2-b^2$ and ${\frac{axsin\theta}{cos^2\theta}} - {\frac{bycos\theta}{sin^2\theta}} = 0$ find the value of $(ax){^\frac{2}{3}} + (by){^\frac{2}{3}}$

Relevant Equations

N/A, can be done through general algebraic techniques.

I was trying this problem from some 2 hours but unable to crack it. I cannot proceed after few lines i.e$${\frac{axsin\theta}{cos^2\theta}} - {\frac{bycos\theta}{sin^2\theta}} = 0$$ $${\frac{axsin\theta}{cos^2\theta}} = {\frac{bycos\theta}{sin^2\theta}}$$ $${\frac{ax}{by}} ={\frac{cos^3\theta}{sin^3\theta}}$$ at last $$cot\theta = {\frac {ax^\frac{1}{3}} {by^\frac{1}{3}}}$$ This is all what I am able to do.

 

[anuttarasammyak]

From your work
by=ax\ tan^3\theta
You can delete $by$ by it.

 

[sahilmm15]

From your work
by=ax\ tan^3\theta
You can delete $by$ by it.

It is looking horrible.

 

[sahilmm15]

It is looking horrible.

Too lengthy I guess, also note I didn't found use of the 1st equation, so maybe we can find something there.

 

[sahilmm15]

This problem is fairly difficult as it was asked in JEE Advanced.

 

[archaic]

we mustn't have $a$ and $b$ in the result?

 

[sahilmm15]

we mustn't have $a$ and $b$ in the result?

We can have $a$ and $b$.

 

[anuttarasammyak]

Try
\frac{ax}{cos\theta}+\frac{by}{sin\theta}=\frac{ax}{cos\theta}[1+\frac{\sin^2\theta}{cos^2\theta}]=...=a^2-b^2
You will get ax as function of a, b and $\theta$. Then
(ax)^{2/3}+(by)^{2/3}=(ax)^{2/3}[ 1 + tan^2\theta]=...

 

[sahilmm15]

Try
\frac{ax}{cos\theta}+\frac{by}{sin\theta}=\frac{ax}{cos\theta}[1+\frac{\sin^2\theta}{cos^2\theta}]=...=a^2-b^2
You will get ax by function of a, b and $\theta$. Then
(ax)^{2/3}+(by)^{2/3}=(ax)^{2/3}[ 1 + tan^2\theta]=...

I am close I guess, need to convert into latex which is another nightmare for me because am just a beginner.

 

[archaic]

We can have $a$ and $b$.

ok, then, perhaps, if you put $A=ax$ and $B=by$, the system will look cleaner and clearer.
$$(1):\,\left(\frac{1}{\cos\theta}\right)A+\left(\frac{1}{\sin\theta}\right)B=a^2-b^2$$$$(2):\,\left(\frac{\sin\theta}{\cos^2\theta}\right)A-\left(\frac{\cos\theta}{\sin^2\theta}\right)B=0$$
you can use the usual tricks to eliminate one of the variables.
for example, if you do $(1)+\frac{\sin\theta}{\cos\theta}(2)$, you will get some expression in $A$ only.

 

[sahilmm15]

After $cot\theta = {\frac {ax^\frac{1}{3}} {by^\frac{1}{3}}}$, let us assume a right angled triangle. From $cot\theta$ we get the base as $ax^\frac{1}{3}$ and perpendicular $by^\frac{1}{3}$. From Pythagoras theorem we get the hypotenuse as $\sqrt{ (ax)^{\frac{2}{3}} + (by)^{\frac {2}{3}} }$. Now we have $${\frac{ax}{cos\theta}} + {\frac{by}{sin\theta}} = a^2-b^2$$. We apply the values of $cos\theta$ and $sin\theta$ from the right angled triangle respectively. After simplification we get $\sqrt{ (ax)^{\frac{2}{3}} + (by)^{\frac {2}{3}} }\cdot [(ax)^{\frac{2}{3}} + (by)^{\frac {2}{3}}]= a^2 - b^2$. Now we know that $\sqrt{x}\cdot x=x^\frac{3}{2}.$. So $$ { [(ax)^{\frac{2}{3}} + (by)^{\frac {2}{3}}]^{\frac{3}{2}} } = a^2-b^2$$. Hence we get $$(ax)^{\frac{2}{3}} + (by)^{\frac {2}{3}} = (a^2-b^2)^{\frac{2}{3}}$$ Difficult question but worth a try.

 

[anuttarasammyak]

You choose the way to delete $\theta$ and I choose the way to delete $by$ to get the same result.