QJEEM25Ma1309: JEE Main 2025 Mathematics - Class 11 JEEM25Ma1309 Solution: Let the parabola , meet the coordinate axes at the points and R …

The given parabola is $y = x^2 + px - 3$ .

Intersection with the y-axis:

At $x = 0$ , we find $y = -3$ .

Thus, the parabola intersects the y-axis at the point $(0, -3)$ .

Circle Equation:

We are given the circle has its center at $(-1, -1)$ and it passes through the points where the parabola intersects the axes. The radius can be found using the distance from the center to any given point the circle passes through. Using $(0, -3)$ :

$\text{Radius} = \sqrt{(0 + 1)^2 + (-3 + 1)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{5}$

Therefore, the equation of the circle is:

$(x + 1)^2 + (y + 1)^2 = 5$

This simplifies to:

$x^2 + 2x + 1 + y^2 + 2y + 1 = 5$

$x^2 + y^2 + 2x + 2y + 2 = 5$

$x^2 + y^2 + 2x + 2y - 3 = 0$

Intersection with the x-axis:

When $y = 0$ , solving the quadratic $x^2 + 2x - 3 = 0$ gives:

$(x + 3)(x - 1) = 0$

So, $x = -3$ or $x = 1$ .

Thus, the intersection points on the x-axis are $(-3, 0)$ and $(1, 0)$ .

Vertices of Triangle $\triangle PQR$ :

The vertices of the triangle formed are $P = (0, -3)$ , $Q = (-3, 0)$ , and $R = (1, 0)$ .

Area of $\triangle PQR$ :

Use the determinant formula to find the area of the triangle:

$\text{Area} = \frac{1}{2} \left| \begin{array}{ccc} 0 & -3 & 1 \\ -3 & 0 & 1 \\ 1 & 0 & 1 \end{array} \right|$

Calculate the determinant:

$= \frac{1}{2} \left( (0)(0)(1) + (-3)(1)(1) + (1)(-3)(1) - (1)(0)(1) - (-3)(1)(0) - (0)(-3)(1) \right)$

Simplify:

$= \frac{1}{2} \left( 0 - 3 + 3 + 0 \right) = \frac{1}{2} (12) = 6$

Thus, the area of $\triangle PQR$ is 6.