JEE Main 2023 (Online) 11th April Morning ShiftQuestion 1
Let $R$ be a rectangle given by the lines $x=0$, $x=2$, $y=0$ and $y=5$. Let $A(\alpha,0)$ and $B(0,\beta)$, $\alpha\in[0,2]$ and $\beta\in[0,5]$, be such that the line segment $AB$ divides the area of the rectangle $R$ in the ratio $4:1$. Then, the mid-point of $AB$ lies on a
Points and Straight Lines
Solution
The rectangle has area $2\cdot5=10$. $AB$ cuts off the right triangle $OAB$ of area $\frac{\alpha\beta}{2}$. <svg fill="none" stroke="#000" font-family="serif" font-size="10" viewBox="0 0 210 347"><path stroke-width=".8" d="M12 295.9h186M51.2 335.1V12"/><path fill="#000" stroke="none" d="m198 295.9-5-2.2v4.4zM51.2 12 49 17h4.4z"/><text x="203.6" y="305.7" fill="#000" stroke="none" font-style="italic">x</text><text x="57.2" y="15.5" fill="#000" stroke="none" font-style="italic">y</text><text x="45.6" y="305.7" fill="#000" stroke="none" font-style="italic" text-anchor="end">O</text><path stroke-width="1.1" d="M51.2 295.9h97.9V51.2H51.2Z"/><path fill="#e6e6e6" d="M51.2 295.9h78.3L51.2 173.5Z"/><path stroke-dasharray="4 3" d="m60.9 51.2 1 22.2 1 18.5 1 15.7 1 13.5.9 11.6 1 10.2 1 9 1 8 1 7.2.9 6.4 1 5.9 1 5.3 1 4.8 1 4.4.9 4.1 1 3.8 1 3.5 1 3.2.9 3 1 2.8 1 2.6 1 2.5 1 2.3.9 2.2 1 2.1 1 1.9 1 1.8 1 1.8.9 1.6 1 1.6 1 1.5 1 1.4 1 1.4.9 1.3 1 1.2 1 1.2 1 1.1.9 1.1 1 1 1 1"/><circle cx="129.5" cy="295.9" r="1.9" fill="#000" stroke-width=".7"/><text x="129.5" y="307.4" fill="#000" stroke="none" text-anchor="middle">A(α,0)</text><circle cx="51.2" cy="173.5" r="1.9" fill="#000" stroke-width=".7"/><text x="44.2" y="177" fill="#000" stroke="none" text-anchor="end">B(0,β)</text><circle cx="90.3" cy="234.7" r="1.9" fill="#000" stroke-width=".7"/><text x="95.9" y="232.6" fill="#000" stroke="none">M</text><text x="156.1" y="84" fill="#000" stroke="none" font-size="8.5">xy=1</text></svg> Since $\alpha\le2$ and $\beta\le5$, this triangle has area at most $5$, so it is the smaller part: $\frac{\alpha\beta}{2}=\frac15\cdot10=2$, i.e. $\alpha\beta=4$. Mid-point $M(h,k)=\left(\frac\alpha2,\frac\beta2\right)$: $hk=\frac{\alpha\beta}{4}=1$. The locus $xy=1$ is a hyperbola. Answer (C).