JEE MAIN - Mathematics (2023 - 12th April Morning Shift - No. 7)
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 :
$$-$$7
7
$$-$$7$$\sqrt3$$
7$$\sqrt3$$
Explanation
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$$ \begin{gathered} x \cos \theta+y \sin \theta=7 \\\\ x-\text { intercept }=\frac{7}{\cos \theta} \\\\ y-\text { intercept }=\frac{7}{\sin \theta} \\\\ \mathrm{A}:\left(\frac{7}{\cos \theta}, 0\right) \mathrm{B}:\left(0, \frac{7}{\sin \theta}\right) \end{gathered} $$
Locus of mid point M : (h, k)
$$ \begin{aligned} & \mathrm{h}=\frac{7}{2 \cos \theta}, \mathrm{k}=\frac{7}{2 \sin \theta} \\\\ & \frac{7}{2 \sin \theta}=\frac{7 \sqrt{3}}{3} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{3} \\\\ & \alpha=\frac{7}{2 \cos \theta}=7 \end{aligned} $$
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