Analytical Geometry!

We know that the equation of the tangent line passing through $(1,1)$ is $y=2x-1$ . And $D$ is the midpoint of $AB$ .

Let $\gamma=\dfrac{CD}{CP}, t_1=\dfrac{CA}{CE}=1+\lambda_{1}, t_2=\dfrac{CB}{CF}=1+\lambda_{2}$ . Then $t_1+t_2=3$ .

We know that $AD=\dfrac{AB}{2}$ , $[CAB]= 2[CAD]=2[CBD]$ . But we have $\dfrac{1}{t_1t_2}=\dfrac{(CE)(CF)}{(CA)(CB)}=\dfrac{[CEF]}{[CAB]}=\dfrac{[CEP]}{2[CAD]}+\dfrac{[CFP]}{2[CED]}=\dfrac{3}{2t_1t_2\gamma}$ .

Thus, $\gamma=\dfrac{3}{2}$ . And $P$ is the centroid of $\triangle ABC$ .

Then consider the points $P(x, y)$ and $C(x_0, x_{0}^{2})$ . Since $C$ is different from $A$ , $x_0 \neq 1.$ And the coordinates of $P$ are $(\dfrac{1+x_0}{3}, \dfrac{x_{0}^{2}}{3})$ . Keeping in mind that $x \neq \dfrac{2}{3}$ . Eliminating $x_0,$ we obtain $y=\dfrac{1}{3}(3x-1)^2$ . So the equation of the trail is $y=\dfrac{1}{3}(3x-1)^2, x\neq \dfrac{2}{3}$ .

Therefore, the answer is $3-1+3=\boxed{5}$ .

A DUMB SOLUTION

Just substitute C $\equiv$ (-1,1) and (0,0) , $\lambda_{1}=1$ and $\lambda_{2}=0$ . You will get two constraints:

• $3b^{2}=a$

• $m=-3b$

Put m=3 (Its clear by $2^{nd}$ constraint that m is a multiple of 3 to find out a and b.