Triangle in parabola I have a problem. In my triangle one vertex is in the vertex of the parabola and two others are in parabola. This is a isosceles triangle and I know one angle in this triangle : 120 grades. The question is: angle 120 grades must be beside of the vertex of parabola?
Thanks a lot for answer, I am beginner in math.

Which option (1) or (2) is possible?
 A: If by 120 you mean $120^\circ $ then it must be on the vertex of parabola. 
Proof is easy. Just take any point A on parabola, which is not vertex (let vertex point be point V) of parabola, then draw line from this point to vertex, then draw line from this point, such that it will form angle with $120^\circ $. You will get point B, which is intersection of new line with other leg of parabola. The two segments AV and AB are not equal,  easy to prove. 
A: choose the coordinates so that the equation of the parabola is $y = x^2.$ let the triangle be $OAB$ with $O =(0,0), A = (a, a^2), B= (b, b^2)$ and $\angle AOB = 120^\circ$ by cosine rule, $$AB^2 = OA^2 + OB^2 + OA*OB $$ so that 
$$(a-b)^2 + (a^2 - b^2)^2 = a^2 + a^4 + b^2 + b^4 + \sqrt{(a^2 + a^4)(b^2+b^4)} $$
cleaning it up a bit we end up with $$3a^2b^2 + 8ab - a^2 - b^2 + 3 = 0$$ 
which simplifies to  $$-2ab(1+ab) =   \sqrt{(a^2 + a^4)(b^2+b^4)} \tag 1$$  from $(1)$ we can see that $a$ and $b$ cannot be of the same sign. using symmetry we can assume that $a > 0, b < 0.$
a special solution is $-b= a = \sqrt{1+\sqrt 2}$ to find the general solution we will square $(1)$ and get a quadratic equation for $b$ $$b^2(3a^2 - 1) + 8ab + (3-a^2) = 0 $$  and the solution is $$ b = \dfrac{-2a \pm \sqrt 3 |a^2 -1|}{3a^2 - 1} \text{ and } b < 0.$$
if you take the negative solution you find that a negative solution exist for $0 \le < \sqrt 3$ and for the positive solution you have $\dfrac{1}{\sqrt 3} < a.$
