**13.** Let the parabola \( y = x^2 + px - 3 \) 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 \( \triangle PQR \) is: - (1) 4 - (2) 6 - (3) 7 - (4) 5

1. Axis-Intercepts of a Parabola

For any parabola of form
$y = f(x) = x^2 + p x - 3$

• y-intercept: put $x = 0$. We get $y = -3 \Rightarrow P(0, -3)$.
• x-intercepts: put $y = 0$. We solve the quadratic
  $x^2 + p x - 3 = 0$
  and obtain two roots $\alpha$ and $\beta$. These give $Q(\alpha, 0)$ and $R(\beta, 0)$.

2. What Does “Passes Through” Mean for a Circle?

If a circle with centre $C(h, k)$ passes through a point $(x_1, y_1)$, the distance between $(h, k)$ and $(x_1, y_1)$ equals the radius $r$. Mathematically,
$(x_1 - h)^2 + (y_1 - k)^2 = r^2$

Here the centre is fixed at $C(-1, -1)$. We compute the distance of each intercept to $C$ and equate them:

1. First get $r$ by using the easiest intercept (usually the y-intercept because we already know its coordinates).
2. Plug the unknown roots $(\alpha, 0)$ and $(\beta, 0)$ into the same distance formula. This pins down their possible values.

3. Area of a Triangle Formed by Three Intercepts

When two vertices lie on the x-axis and the third lies on the y-axis (or any vertical line), the triangle is right-angled at the origin’s projection. Using the simple formula
$\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height}$ with

• base = horizontal distance between the two x-intercepts
• height = vertical distance of the y-intercept from the x-axis makes life easy and avoids the longer Shoelace/Determinant method.

What’s happening here?

Imagine you draw a smiling U-shaped curve (that’s the parabola) on your graph paper.
It cuts the axes in three places:

• once on the y-axis (straight up–down line)
• twice on the x-axis (left–right line)

Now someone puts a round sticker (a circle) with its centre fixed at the point (−1, −1). The sticker just touches those three cutting points. We simply need the area of the triangle formed by those three points.

To tackle it we:

1. Find the three cutting points (called intercepts).
2. Force them to sit on the circle (distance from centre must be the same). This locks the two x-intercepts to particular values.
3. Finally, use the plain old ½ × base × height rule to get the area. That’s all!

Step 1 – Identify Intercepts

• y-intercept (P): set $x = 0$
  $y = 0^2 + p(0) - 3 = -3 \implies P(0, -3)$

• x-intercepts (Q & R): set $y = 0$
  $x^2 + p x - 3 = 0 \quad (1)$ Let roots be $\alpha, \beta$. So
  $Q(\alpha, 0), \quad R(\beta, 0)$

Step 2 – Use the Circle Condition

Circle centre $C(-1, -1)$.

1. Radius via P: $CP^2 = (0 + 1)^2 + (-3 + 1)^2 = 1^2 + (-2)^2 = 1 + 4 = 5$ Hence $r^2 = 5$.

2. Enforce same radius for Q (and R): $CQ^2 = (\alpha + 1)^2 + (0 + 1)^2 = (\alpha + 1)^2 + 1 = 5$ $(\alpha + 1)^2 = 4 \implies \alpha + 1 = \pm 2$ $\boxed{\alpha = 1 \;\text{or}\; \alpha = -3}$ The same will hold for $\beta$. Therefore the two roots are $\{1, -3\}$.

Step 3 – Recover p

Using sum and product of roots for equation (1):

$\alpha + \beta = -p, \qquad \alpha \beta = -3$

Here $\alpha + \beta = 1 + (-3) = -2 \;\Rightarrow\; -p = -2 \implies p = 2$ (consistent with $\alpha \beta = -3$).

Step 4 – Co-ordinates of Q & R

$Q(1, 0), \qquad R(-3, 0)$

Step 5 – Area of \triangle PQR

Base $QR$ lies on x-axis:

$\text{base} = |1 - (-3)| = 4$

Height is the $y$-distance of $P$ from x-axis:

$\text{height} = |-3| = 3$

Therefore

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

[ \boxed{6}\ ]
Option (2) is correct.