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. 5
• D. 7

Answer 1

Answer: (B)

6

Answer 2

The parabola is given by

$$y=x^2+px-3.$$

It meets the y-axis when $x=0$, so one point is

$$P=(0,-3).$$

The x-intercepts are found by setting $y=0$:

$$x^2+px-3=0.$$

Let the roots be $x_1$ and $x_2$. Then the x-axis intersection points are

$$Q=(x_1,0) \quad \text{and} \quad R=(x_2,0).$$

A circle $C$ with center $(-1,-1)$ passes through all three points. Its radius using point $P$ is:

$$r=\sqrt{(0+1)^2+(-3+1)^2} = \sqrt{1+4}=\sqrt{5}.$$

Hence, for a point $(x, y)$ on the circle,

$$(x+1)^2+(y+1)^2=5.$$

For the x-intercepts $Q=(x_1,0)$ and $R=(x_2,0)$:

$$(x_1+1)^2+(0+1)^2 = 5 \quad \Rightarrow \quad (x_1+1)^2+1=5,$$ $$(x_1+1)^2=4 \quad \Rightarrow \quad x_1+1=\pm2.$$

Thus, $x_1=1$ or $x_1=-3$. Hence, the roots of the quadratic are $1$ and $-3$.

Using Vieta’s formulas for $x^2+px-3=0$:

• Sum of roots: $1+(-3)=-2=-p \Rightarrow p=2$.
• Product of roots: $1\times(-3)=-3$ (consistent with the constant term).

The vertices of $\triangle PQR$ are:

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

Calculate the area using the base and height method. The base along the x-axis from $(-3,0)$ to $(1,0)$ has length:

$$(1 - (-3))=4.$$

The height from point $(0,-3)$ to the x-axis is $3$.

Thus, the area $A$ is:

$$A = \frac{1}{2}\times\text{base}\times\text{height} = \frac{1}{2}\times 4\times 3 = 6.$$