If x=acos3θsin2θ,y=asin3θcos2θand (x2+y2)p(xy)q,(p,q∈N)is independent of θ then

Given that,

x = a \cos^{3} θ \sin^{2} θ

y = a \sin^{3} θ \cos^{2} θ

By substituting the above values in

{(x^{2} + y^{2})}^{p}

, we get

= {(a^{2} \cos^{6} θ \sin^{4} θ + a^{2} \sin^{6} θ \cos^{4} θ)}^{p}

= {[a^{2} \sin^{4} θ \cos^{4} θ (\cos^{2} θ + \sin^{2} θ)]}^{p}

= (a^{2} \sin^{4} θ \cos^{4} θ)^{p}

Also, by substituting in

{(xy)}^{q}

, we get

= {(a^{2} \cos^{5} θ \sin^{5} θ)}^{q}

Now let us find,

\frac{{(x^{2} + y^{2})}^{p}}{{(xy)}^{q}}

= \frac{{(a^{2} \sin^{4} θ \cos^{4} θ)}^{p}}{(a^{2} \cos^{5} θ \sin^{5} θ)^{q}}

= a^{2 p - 2 q} \times (sinθcosθ)^{4 p - 5 q}

Now from the above expression we can observe,

\frac{{(x^{2} + y^{2})}^{p}}{{(xy)}^{q}}

is independent of

θ

,
if

4 p - 5 q = 0

∴ 4 p = 5 q

Hence, the correct option is (2).

If $x = a \cos^{3} θ \sin^{2} θ, y = a \sin^{3} θ \cos^{2} θ$ and $\frac{(x^{2} + y^{2})^{p}}{(xy)^{q}}, (p, q \in N)$ is independent of $θ$ then