If the curves $$\frac{x^2}{\alpha} + \frac{y^2}{4} = 1$$ and $$y^3 = 16x$$ intersect at right angles, then a value of $$\alpha$$ is :

Differentiating the ellipse $$\frac{x^{2}}{\alpha} + \frac{y^{2}}{4} = 1$$ implicitly gives $$\frac{2x}{\alpha} + \frac{2y}{4}\,\frac{dy}{dx} = 0$$, so $$\frac{dy}{dx}\bigg|_{\text{ellipse}} = -\frac{4x}{\alpha y}$$.

Differentiating the curve $$y^{3} = 16x$$ gives $$3y^{2}\,\frac{dy}{dx} = 16$$, so $$\frac{dy}{dx}\bigg|_{\text{curve}} = \frac{16}{3y^{2}}$$.

For the two curves to intersect at right angles, the product of their slopes at the point of intersection must equal $$-1$$: $$\left(-\frac{4x}{\alpha y}\right)\!\left(\frac{16}{3y^{2}}\right) = -1$$. This simplifies to $$\frac{64x}{3\alpha y^{3}} = 1$$.

From the second curve, $$y^{3} = 16x$$, so substituting gives $$\frac{64x}{3\alpha \cdot 16x} = 1$$, which reduces to $$\frac{4}{3\alpha} = 1$$, hence $$\alpha = \dfrac{4}{3}$$.