JEE Main 2023 (Online) 12th April Morning ShiftQuestion 1
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
Points and Straight Lines
Solution
Intercepts $\frac7{\cos\theta},\frac7{\sin\theta}$; midpoint $(h,k)=\left(\frac{7}{2\cos\theta},\frac{7}{2\sin\theta}\right)$. $\frac{49}{4h^2}+\frac{49}{4k^2}=1\Rightarrow\frac1{h^2}+\frac1{k^2}=\frac4{49}$. $k=\frac{7\sqrt3}{3}$: $\frac1{k^2}=\frac3{49}\Rightarrow h^2=49$; $\theta\in(0,\frac\pi2)\Rightarrow h>0$, $\alpha=7$. Option (C).