Question

If $\frac{ax}{\cos \theta }+\frac{by}{\sin \theta }=(a^2-b^2)$ , and $\frac{ax\sin \theta }{{\cos }^2\theta }-\frac{by\cos \theta }{{\sin }^2\theta }=0$, then $(ax{)}^{2/3}+(by{)}^{2/3}=$

• A. $(a^2-b^2{)}^{2/3}$
• B. $(a^2+b^2{)}^{2/3}$
• C. $(a^2+b^2{)}^{3/2}$
• D. $(a^2-b^2{)}^{3/2}$

Answer 1

The correct option is A $(a^2-b^2{)}^{2/3}$

Let $\frac{1}{cos\theta }=l$ and $\frac{1}{sin\theta }=m$

The above expressions can be re-written as

$ax\left(l\right)+by\left(m\right)=(a^2-b^2)$ ...(i)
$ax(\frac{l^2}{m})-by(\frac{m^2}{l})=0$ ..(ii)

Hence,

$\left(axl\right)=\frac{by\left(m^3\right)}{l^2}$

Substituting in (i) we get

$by\left(m\right)(\frac{m^2}{l^2}+1)=(a^2-b^2)$

$bym(\frac{m^2+l^2}{l^2})=(a^2-b^2)$

Now,
$\frac{1}{cos\theta }=l$ and $\frac{1}{sin\theta }=m$

Hence,

$\frac{1}{l^2}+\frac{1}{m^2}=1$

$l^2+m^2=l^2{m}^2$

Substituting we get.

$by\left(m\right)(\frac{l^2{m}^2}{l^2})=(a^2-b^2)$

$by\left(m^3\right)=(a^2-b^2)$

$sin\theta =(\frac{by}{a^2-b^2}{)}^{1/3}$ ...(iii)

Similarly we can get

$cos\theta =(\frac{ax}{a^2-b^2}{)}^{1/3}$ ...(iv)

Squaring and adding (iii) and (iv) we get

$(\frac{by}{a^2-b^2}{)}^{2/3}+(\frac{ax}{a^2-b^2}{)}^{2/3}=1$

$(by{)}^{2/3}+(ax{)}^{2/3}=(a^2-b^2{)}^{2/3}$