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

To find the area of \(\triangle PQR\), we begin by determining the coordinates of points P, Q, and R where the parabola \(y = x^2 + px - 3\) intersects the axes.

Step 1: Find the x-intercepts (roots of the parabola).
Set \(y = 0\) to solve for \(x\):
\(x^2 + px - 3 = 0\).
The solutions are \(x_1\) and \(x_2\), giving us P(\(x_1\), 0) and Q(\(x_2\), 0).

Step 2: Find the y-intercept (where the parabola meets the y-axis).
Set \(x = 0\):
\(y = -3\).
This gives us the point R(0, -3).

Step 3: Find circle C's equation using center (-1, -1). It passes through P, Q, and R.
Equation: \((x + 1)^2 + (y + 1)^2 = r^2\).

Step 4: Substitute R(0, -3) to find \(r\):
\((0 + 1)^2 + (-3 + 1)^2 = r^2\),
\(1 + 4 = r^2\),
\(r^2 = 5\).

Step 5: Equation of circle becomes:
\((x + 1)^2 + (y + 1)^2 = 5\).

Substitute P and Q coordinates, obtain \(x_1\) and \(x_2\):
Step 6: Solve:\(x_1 + x_2 = -p\) (sum of roots formula for quadratic),
\(x_1x_2 = -3\) (product of roots formula for quadratic).
Step 7: Using circle property, substitute P(x1,0):
\((x_1 + 1)^2 + 1 = 5\),
\(x_1^2 + 2x_1 + 1 + 1 = 5\),
\(x_1^2 + 2x_1 - 3 = 0\),
Similarly for Q, get \(x_2^2 + 2x_2 - 3 = 0\).
Step 8: Calculate area of \(\triangle PQR\).
Coordinates: P(\(x_1\),0), Q(\(x_2\),0), R(0,-3).

Area	=	\(\frac{1}{2}|x_1(0 + 3) + x_2(-3 - 0) + 0(0 - 0)|\)
	=	\(\frac{1}{2}|3x_1 - 3x_2|\)
	=	\(\frac{3}{2}|x_1 - x_2|\)

Recall: \(x_1,x_2\) roots of \(x^2 + 2x - 3 = 0\).
Since \((x_1-x_2)^2=(x_1+x_2)^2 - 4x_1x_2\), we calculate the difference:
\((x_1-x_2)^2 = 4 - 4(-3) = 16\),
\(x_1-x_2 = \pm 4\),
So, \(|x_1-x_2| = 4\).

Area = \(\frac{3}{2}\times 4\) = 6.

The area of \(\triangle PQR\) is 6.

Answer 2

The intersections with the $x$–axis are the roots of $x^2+px-3=0$. Let these roots be $r_1,r_2$, so the $x$–intercepts are $P(r_1,0)$ and $Q(r_2,0)$.

The $y$–intercept (at $x=0$) is $R(0,-3)$.

Distance from the circle centre $(-1,-1)$ to $R(0,-3)$: $$ \text{radius}^2 = (0+1)^2 + (-3+1)^2 = 1 + 4 = 5. $$ So every point on the circle satisfies $(x+1)^2+(y+1)^2=5$.

For the $x$–intercepts $P(r,0)$ the circle equation gives $$ (r+1)^2 + (0+1)^2 = 5 \quad\Rightarrow\quad (r+1)^2+1=5 \Rightarrow (r+1)^2=4. $$ Hence $r+1=\pm 2$, so the two roots are $r=1$ and $r=-3$.

Thus the three vertices are $$P(1,0),\qquad Q(-3,0),\qquad R(0,-3).$$

The base $PQ$ has length $|1-(-3)|=4$ and the height from $R$ to the $x$-axis is $3$. So the area is $$ \text{Area}=\tfrac{1}{2}\times\text{base}\times\text{height}=\tfrac{1}{2}\times 4\times 3 = 6. $$

Answer

Area $=6$. (Option 2)