isi-entrance 2023 Q26

isi-entrance · India · UGA Straight Lines & Coordinate Geometry Locus Determination
The straight line $O A$ lies in the second quadrant of the $(x, y)$-plane and makes an angle $\theta$ with the negative half of the $x$-axis, where $0 < \theta < \frac { \pi } { 2 }$. The line segment $C D$ of length 1 slides on the $(x, y)$-plane in such a way that $C$ is always on $O A$ and $D$ on the positive side of the $x$-axis. The locus of the mid-point of $C D$ is
(A) $x ^ { 2 } + 4 x y \cot \theta + y ^ { 2 } \left( 1 + 4 \cot ^ { 2 } \theta \right) = \frac { 1 } { 4 }$.
(B) $x ^ { 2 } + y ^ { 2 } = \frac { 1 } { 4 } + \cot ^ { 2 } \theta$.
(C) $x ^ { 2 } + 4 x y \cot \theta + y ^ { 2 } = \frac { 1 } { 4 }$.
(D) $x ^ { 2 } + y ^ { 2 } \left( 1 + 4 \cot ^ { 2 } \theta \right) = \frac { 1 } { 4 }$.
The straight line $O A$ lies in the second quadrant of the $(x, y)$-plane and makes an angle $\theta$ with the negative half of the $x$-axis, where $0 < \theta < \frac { \pi } { 2 }$. The line segment $C D$ of length 1 slides on the $(x, y)$-plane in such a way that $C$ is always on $O A$ and $D$ on the positive side of the $x$-axis. The locus of the mid-point of $C D$ is\\
(A) $x ^ { 2 } + 4 x y \cot \theta + y ^ { 2 } \left( 1 + 4 \cot ^ { 2 } \theta \right) = \frac { 1 } { 4 }$.\\
(B) $x ^ { 2 } + y ^ { 2 } = \frac { 1 } { 4 } + \cot ^ { 2 } \theta$.\\
(C) $x ^ { 2 } + 4 x y \cot \theta + y ^ { 2 } = \frac { 1 } { 4 }$.\\
(D) $x ^ { 2 } + y ^ { 2 } \left( 1 + 4 \cot ^ { 2 } \theta \right) = \frac { 1 } { 4 }$.