- Home
- Standard 11
- Mathematics
If the point $\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $x$ $\cos \theta+ y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)$ between the coordinates axes, then $\alpha$ is equal to
$7$
$-7$
$-7 \sqrt{3}$
$7 \sqrt{3}$
Solution
$\operatorname{pt}\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$
$x -\text { intercept }=\frac{7}{\cos \theta}$
$y -\text { intercept }=\frac{7}{\sin \theta}$
$A:\left(\frac{7}{\cos \theta}, 0\right) B :\left(0, \frac{7}{\sin \theta}\right)$
Locus of mid pt $M 🙁 h , k )$
$h =\frac{7}{2 \cos \theta}, k =\frac{7}{2 \sin \theta}$
$\frac{7}{2 \sin \theta}=\frac{7 \sqrt{3}}{3} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{3}$
$\alpha=\frac{7}{2 \cos \theta}=7$
