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QJEEM23Ma1c68: JEE Main 2023 Mathematics - Class 11 JEEM23Ma1c68 Solution: Let R be a rectangle given by the lines and . Let A and B and , …

Let R be a rectangle given by the lines x=0,x=2,y=0x=0, x=2, y=0 and y=5y=5 . Let A (α,0)(\alpha,0) and B (0,β),α∈[0,2](0,\beta),\alpha\in[0,2] and β∈[0,5]\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 :

Multiple Choice Options

  • A hyperbola
  • B straight line
  • C parabola
  • D circle

Step-by-Step Verified Solution

We have, RR be a rectangle formed by the lines x=0x=0 , x=2,y=0x=2, y=0 and y=5y=5

JEE Main 2023 (Online) 11th April Morning Shift Mathematics - Hyperbola Question 16 English Explanation

Let PQRSP Q R S be the rectangle such that

P(0,0),Q(2,0),R(2,5) and S(0,5)P(0,0), Q(2,0), R(2,5) \text { and } S(0,5)

PQ=SR=2P Q=S R=2 units and PS=QR=5P S=Q R=5 units

∴\therefore Area of rectangle PQRS=(2×5)sqP Q R S=(2 \times 5) \mathrm{sq} units =10sq=10 \mathrm{sq} units

Area of △PAB=15×\triangle P A B=\frac{1}{5} \times Area of rectangle

12×α×βamp;=15×10 sq units amp;=2 sq units 12×α×βamp;=2⇒αβamp;=4 sq units .........(i)\begin{aligned}\frac{1}{2} \times \alpha \times \beta & =\frac{1}{5} \times 10 \text { sq units } \\\\& =2 \text { sq units } \\\\\frac{1}{2} \times \alpha \times \beta & =2 \\\\\Rightarrow \alpha \beta & =4 \text { sq units } .........(i)\end{aligned}

Let M(h,k)M(h, k) be the mid-point of ABA B .

Here, h=α2,k=β2h=\frac{\alpha}{2}, k=\frac{\beta}{2}

or α=2h\alpha=2 h and β=2k\beta=2 k

On substitute values of α\alpha and β\beta in Eq. (i), we get

(2h)(2k)amp;=4⇒hkamp;=1∴ Locus of M is xyamp;=1, which is a hyperbola. \begin{aligned}\quad(2 h)(2 k) & =4 \\\\\Rightarrow h k & =1 \\\\\therefore \text { Locus of } M \text { is } x y & =1 \text {, which is a hyperbola. }\end{aligned}

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