JEE MAIN - Mathematics (2023 - 11th April Morning Shift - No. 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 :
hyperbola
straight line
parabola
circle

Explanation

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

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

Let $P Q R S$ be the rectangle such that

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

$P Q=S R=2$ units and $P S=Q R=5$ units

$\therefore$ Area of rectangle $P Q R S=(2 \times 5) \mathrm{sq}$ units $=10 \mathrm{sq}$ units

Area of $\triangle P A B=\frac{1}{5} \times$ Area of rectangle

$$ \begin{aligned} \frac{1}{2} \times \alpha \times \beta & =\frac{1}{5} \times 10 \text { sq units } \\\\ & =2 \text { sq units } \\\\ \frac{1}{2} \times \alpha \times \beta & =2 \\\\ \Rightarrow \alpha \beta & =4 \text { sq units } .........(i) \end{aligned} $$

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

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

or $\alpha=2 h$ and $\beta=2 k$

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

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

Comments (0)

Advertisement