Let the parabola y=x^2+p x-3 , meet the coordinate axes at the points P, Q and R . If the circle C with centre at (-1,-1) passes through the points P, Q and R , then the area of \triangle P Q R is :

Parabola PYQ — JEE Main 2025 · 22 Jan · Shift 1 · Q41 with Solution

Find the intercepts of the parabola

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

It meets the y-axis at $x=0$ , so $R=(0,-3).$

It meets the x-axis where $y=0$ : $x^2+px-3=0.$ Let the roots be $\alpha,\beta$ . Then the x-intercepts are $P=(\alpha,0),\qquad Q=(\beta,0).$

Use the circle condition

The circle has centre $(-1,-1)$ and passes through $P,Q,R$ . Hence the distances of these three points from $(-1,-1)$ are equal.

First, for $R=(0,-3)$ : $CR^2=(0+1)^2+(-3+1)^2=1^2+(-2)^2=5.$ So the radius satisfies $r^2=5.$

Therefore, $P$ and $Q$ must also satisfy

\quad CQ^2=5.$$ For a general x-intercept $(x,0)$, $$C(x,0)^2=(x+1)^2+(0+1)^2=(x+1)^2+1.$$ Setting this equal to $5$: $$ (x+1)^2+1=5$$ $$ (x+1)^2=4$$ $$ x+1=\pm 2$$ $$ x=1 \text{ or } x=-3.$$ Hence, $$P=(1,0),\qquad Q=(-3,0)$$ (order may be interchanged).

Check consistency with the parabola

Since the x-intercepts are $1$ and $-3$ , the quadratic is $y=(x-1)(x+3)=x^2+2x-3,$ so indeed some value of $p$ exists, namely $p=2$ .

Find the area of $\triangle PQR$

Points are $P=(1,0),\quad Q=(-3,0),\quad R=(0,-3).$

Segment $PQ$ lies on the x-axis, so its length is $PQ=|1-(-3)|=4.$

The perpendicular distance of $R$ from the x-axis is $3.$

Therefore, $\text{Area} = \frac{1}{2}\times 4 \times 3 = 6.$

Compare with stored answer

Our derived answer is 6, which corresponds to Option B. This matches the stored correct answer.