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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 co-ordinates axes, then $\alpha$ is equal to

Option: 1

7\sqrt{3}


Option: 2

-7


Option: 3

7


Option: 4

-7\sqrt{3}


Answers (1)

best_answer

\left(\alpha, \frac{7 \sqrt{3}}{3}\right)

$$ \begin{aligned} & x \cos \theta+y \sin \theta=7 \\ & 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) \end{aligned} \\ \text{ Locus of mid pt 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}

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Ritika Jonwal

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