Geometry -- 3

Let $AB$ be the base of $\triangle ABC$ , point $C$ be $(a,a^2)$ and $CN$ be the altitude from $C$ to $AB$ . Then the equation of $AB$ is $y = 2-x \ ...(1)$ and the equation of $CN$ is given by $\dfrac {y-a^2}{x-a} = 1$ $\implies y = x - a + a^2 \ ...(2)$ .

The coordinates of $N(x_N, y_N)$ are given by $(1) = (2): \ 2-x_N = x_N - a + a^2$ $\implies x_N = \dfrac {2+a-a^2}2$ and from $(1): \ y_N = \dfrac {2-a+a^2}2$ .

The height of $\triangle ABC$ , $CN = \sqrt{(x_N-a)^2+(y_N-a^2)^2} = \sqrt{\left(\dfrac {2-a-a^2}2\right)^2 + \left(\dfrac {2-a-a^2}2\right)^2} = \dfrac {|2-a-a^2|}{\sqrt 2}$ .

The area of $\triangle ABC$ , $[ABC] = \dfrac 12 \times AB \times CN = \dfrac 12 \times 2 \sqrt 2 \times \dfrac {|2-a-a^2|}{\sqrt 2} = |2-a-a^2|$ .

For $[ABC]=2$ ,

$|2-a-a^2| = 2 \implies \begin{cases} {\small \text{For }} 2-a-a^2 > 0 & \implies 2-a-a^2 = 2 & \implies a = -1, 0 & \small \text{2 solutions} \\ {\small \text{For }} 2-a-a^2 < 0 & \implies a^2+a-2 = 2 & \implies a = \dfrac {\pm \sqrt{17} -1}2 & \small \text{2 solutions} \end{cases}$

Therefore, there are $\boxed 4$ possible point $C$ 's.

$|AB| = 2\sqrt 2$ $\\$ $l_{AB}: y=-x+2 --> x+y-2=0$ $\\$ Let $C$ be $(x_o, y_o)$ $\\$ $\\$ $d = \frac {|x_o+y_o-2|}{\sqrt 2 }$ $\\$ $\frac {2\sqrt 2}{2} \cdot \frac {|x_o+y_o-2|}{\sqrt 2 } = |x_o+y_o-2|=2$ $\\$ $\because$ $C$ is on the function $y=x^2$ $\\$ $\therefore y_o=x_o^2$ $\\$ $|x_o+x_o^2-2|=2$ $\\$ when $x_o+x_o^2-2=2$ $\\$ $x_o^2+x_o-4=0$ $\\$ $\Delta = b^2-4ac = 1+16 > 0$ $\boxed {2 \: solutions}$ $\\$ $\\$ when $x_o+x_o^2-2=-2$ $\\$ $x_o^2+x_o=0$ $\\$ $x_o(x_o+1)=0$ $\boxed {2 \: solutions}$ $\\$

Thus $4$ solutions in total