JEE 2025 — Mathematics PYQ

$$y=x^2+px-3$$
$$\because x=0y=-3$$
$$\therefore (0,-3)\text{ be point where parabola (cuts y axis)}$$
$$\text{Equation of circle }{C}^{\prime }$$
$$(x+1{)}^2+(y+1{)}^2=(-1-0{)}^2+(-1+3{)}^2$$
$$\Rightarrow (x+1{)}^2+(y+1{)}^2=5$$

Circle cuts x-axis at y = 0
∴ (x+1)² = 4 → x + 1 = ± 2 → x = 1, -3
∴ Points are (1, 0), (-3, 0) and (0, -3)
Area of Δ = $\frac{1}{2}$×4×3 = 6