Question

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:

• A. 4
• B. 6
• C. 7
• D. 5

Hint:

To find the area of a triangle with vertices on the coordinate axes, use the formula \(\frac{1}{2} \times \text{base} \times \text{height}\).

Answer 1

The Correct Option is B

Step 1: Define the points P, Q, and R.
The parabola is given by \(y = x^2 + px - 3\).
Let P and Q be the points where the parabola intersects the x-axis, so P(\(\alpha\), 0) and Q(\(\beta\), 0).
Let R be the point where the parabola intersects the y-axis, so R(0, -3).

Step 2: Find the equation of the circle.
The circle C has its centre at (-1, -1) and passes through R(0, -3).
The equation of the circle is \((x + 1)^2 + (y + 1)^2 = r^2\).
Since R(0, -3) lies on the circle, we have:
\((0 + 1)^2 + (-3 + 1)^2 = r^2\)
\(1^2 + (-2)^2 = r^2\)
\(1 + 4 = r^2\)
\(r^2 = 5\)
Thus, the equation of the circle is \((x + 1)^2 + (y + 1)^2 = 5\).

Step 3: Find the points P and Q.
Since P and Q lie on the x-axis, their y-coordinates are 0.
Substitute y = 0 into the equation of the circle:
\((x + 1)^2 + (0 + 1)^2 = 5\)
\((x + 1)^2 + 1 = 5\)
\((x + 1)^2 = 4\)
\(x + 1 = \pm 2\)
\(x = -1 \pm 2\)
So, \(x = 1\) or \(x = -3\).
Therefore, P(1, 0) and Q(-3, 0).

Step 4: Calculate the area of \(\triangle PQR\).
The base of \(\triangle PQR\) is PQ, and its length is \(|1 - (-3)| = 4\).
The height of \(\triangle PQR\) is the absolute value of the y-coordinate of R, which is \(|-3|\) = 3.
The area of \(\triangle PQR\) is \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3 = 6\).