If the point ( ,7√33) lies on the curve traced by the mid-points of the line segments of t…

Question

If the point $(\alpha ,\frac{7\sqrt{3}}{3})$ lies on the curve traced by the mid-points of the line segments of the lines $x\cos 	heta +y\sin 	heta =7,	heta \in (0,\frac{\pi }{2})$ between the co-ordinates axes, then $\alpha $ is equal to :

Accepted Answer

$$
\begin{gathered}
x \cos \theta+y \sin \theta=7 \\
x-\text { intercept }=\frac{7}{\cos \theta} \\
y-\text { intercept }=\frac{7}{\sin \theta} \\
\mathrm{A}:\left(\frac{7}{\cos \theta}, 0\right) \mathrm{B}:\left(0, \frac{7}{\sin \theta}\right)
\end{gathered}
$$

Locus of mid point M : (h, k)

$$
\begin{aligned}
& \mathrm{h}=\frac{7}{2 \cos \theta}, \mathrm{k}=\frac{7}{2 \sin \theta} \\
& \frac{7}{2 \sin \theta}=\frac{7 \sqrt{3}}{3} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{3} \\
& \alpha=\frac{7}{2 \cos \theta}=7
\end{aligned}
$$

Solutions · 1