If the point $\left(\mathrm{\alpha },\frac{{\displaystyle 7\sqrt{3}}}{{\displaystyle 3}}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $\mathrm{xcos\theta }+\mathrm{ysin\theta }=7,\mathrm{\theta }\in (0,\frac{\mathrm{\pi }}{2})$ between the co-ordinates axes, then $\mathrm{\alpha }$ is equal to

Given $\mathrm{xcos\theta }+\mathrm{ysin\theta }=7$
$$\begin{gathered} \mathrm{Let} \left({\mathrm{x}}_1,{\mathrm{y}}_1\right) \mathrm{be} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{the} \mathrm{line} \mathrm{segment}. \\ {\mathrm{x}}_1=\frac{7}{2\mathrm{cos\theta }};{\mathrm{y}}_1=\frac{7}{2\mathrm{sin\theta }} \\ {\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }=1 \\ \Rightarrow \frac{49}{4{{\mathrm{x}}_1}^2}+\frac{49}{4{{\mathrm{y}}_1}^2}=1 \end{gathered}$$
$\therefore$ Equation to the locus of  P is $\frac{1}{{\mathrm{x}}^2}+\frac{1}{{\mathrm{y}}^2}=\frac{4}{49}$
Now, $(\mathrm{\alpha },\frac{7\sqrt{3}}{3})$ lies on it $\Rightarrow \frac{1}{{\mathrm{\alpha }}^2}+\frac{3}{49}=\frac{4}{49}$
$\Rightarrow {\mathrm{\alpha }}^2=49\Rightarrow \mathrm{\alpha }=7$