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 $\Delta PQR$ is:
- (1) 4
- (2) 6
- (3) 7
- (4) 5
Solution:
Equation of parabola: $y = x^2 + px – 3$.
It meets the y-axis at point R. Putting $x = 0$, we get $y = -3$.
So, coordinates of R are $(0, -3)$
Circle C has centre $(-1, -1)$ and passes through R $(0, -3)$.
Radius squared $r^2 = (0 – (-1))^2 + (-3 – (-1))^2$
$$ r^2 = (1)^2 + (-2)^2 = 1 + 4 = 5 $$
Equation of circle:
$$ (x+1)^2 + (y+1)^2 = 5 $$
The circle meets the x-axis at points P and Q. Putting $y = 0$ in the circle’s equation:
$$ (x+1)^2 + (0+1)^2 = 5 $$
$$ (x+1)^2 + 1 = 5 \Rightarrow (x+1)^2 = 4 $$
$$ x+1 = \pm 2 \Rightarrow x = 1 \text{ or } x = -3 $$
So, coordinates are $P(1, 0)$ and $Q(-3, 0)$ (or vice versa).
Area of $\Delta PQR$ with vertices $(1, 0)$, $(-3, 0)$ and $(0, -3)$:
$$ \text{Area} = \frac{1}{2} \left| \begin{matrix} 1 & 0 & 1 \\ -3 & 0 & 1 \\ 0 & -3 & 1 \end{matrix} \right| $$
$$ = \frac{1}{2} |1(0 – (-3)) – 0 + 1(9 – 0)| $$
$$ = \frac{1}{2} |3 + 9| = \frac{12}{2} = 6 $$
Ans. (2)