If the point (α, 7 3 3 ) lies on the curve traced by the mid-points...

The straight line is shown below with the intercepts.

$M\left(\dfrac{7}{2cos\theta}, \dfrac{7}{2sin\theta}\right) ≡ \left(h, k\right)$

$∴ cos\theta = \dfrac{7}{2h}sin\theta = \dfrac{7}{2k}$

$∵ \theta \in \left(0, \dfrac{\pi}{2}\right)$

$⇒ \dfrac{49}{4{h}^{2}} + \dfrac{49}{4{k}^{2}} = 1 ∴ x$ and $y$ are positive

$∴$ Locus $= \dfrac{1}{{x}^{2}} + \dfrac{1}{{y}^{2}} = \dfrac{4}{49}$

Now $\left(\alpha, \dfrac{7\sqrt{3}}{3}\right)$ lies on curve

$⇒ \dfrac{1}{{\alpha}^{2}} + \dfrac{9}{147} = \dfrac{4}{49}$

$\dfrac{1}{{\alpha}^{2}} = \dfrac{1}{49}$

$∴ \alpha = \pm 7$

$∴ \alpha = 7$

Hence this is the required option.