If the tangent at θ = π/4 to the curve $x=a\cos^3θ,y=a\sin^3θ$ meets the x and y axis in A and B, then the area of ΔOAB is:

$\frac{a^2}{4}$

$\frac{dy}{dx}=\frac{3a\sin θ\cos θ}{-3a\cos^2 θ\sin θ}=-\tan θ⇒(\frac{dy}{dx})_{θ = π/4}=-1$

So, Point = $\frac{a}{2\sqrt{2}},\frac{a}{2\sqrt{2}}$

Equation of the tangent is $y-\frac{a}{2\sqrt{2}}=-1(x-\frac{a}{2\sqrt{2}})$

$⇒x+y=\frac{a}{\sqrt{2}}$

∴ Area of ΔOAB = $\frac{\frac{a^2}{2}}{2|1×1|}=\frac{a^2}{4}$