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I have this parametric plot: $$x=A{t^2}+{V_0}\cos(\theta){t}\\ y=B{t^2}+{V_0}\sin(\theta){t}$$ which $A$ and $B$ are constant. I want to find $V_0$ and $\theta$ when plot crosses points $(x1,y1)$ and $(x2,y2)$.

Means i have a 4 equations 4 unknowns system: $$x_1=A{t_1^2}+{V_0}\cos(\theta){t_1}\\ x_2=A{t_2^2}+{V_0}\cos(\theta){t_2}\\ y_1=B{t_1^2}+{V_0}\sin(\theta){t_1}\\ y_2=B{t_2^2}+{V_0}\sin(\theta){t_2}$$

Now i have serious problem solving this system for $V_0$ and $\theta$. Please help me

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    With $$\frac{y-Bt^2}{x-At^2}=\tan\theta$$ you will only have one unknown. Can you continue from there? – 2017-02-05
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    now i have 1 equation and 2 unknowns: $t$ and $\theta$ for points $(x_1,y_1)$ and $(x_2,y_2)$ i have: $$\frac{y_1-Bt_1^2}{x_1-At_1^2}=tan\theta$$ $$\frac{y_2-Bt_2^2}{x_1-At_2^2}=tan\theta$$ which means now i have 2 equations and 3 unknowns. – 2017-02-05
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    If it really is a projectile motion, $A$ is supposed to be $0$ – 2017-02-05
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    @N74 $A$ simulate wind and it is $0$ when there is no wind that system can be solved easily. here the path is not a parabola because of wind. – 2017-02-05
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    $t_1$ and $t_2$ are also unknowns, so you need at least another condition to solve the system (assuming $t_1=0$) – 2017-02-05
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    @N74 $t_1$ and $t_2$ are related to points $(x_1,y_1)$ and $(x_2,y_2)$. with $\theta$ and $V_0$ we have 4 equations and 4 unknowns so no more conditions needed. the problem is solving that system. – 2017-02-05
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    In your equations above $x_1=x_2$ and $y_1=y_2$. I assume this is a mistake? – 2017-02-06

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Starting from $$x_1=A{t_1^2}+{V_0}\cos(\theta){t_1}\\ x_2=A{t_2^2}+{V_0}\cos(\theta){t_2}\\ y_1=B{t_1^2}+{V_0}\sin(\theta){t_1}\\ y_2=B{t_2^2}+{V_0}\sin(\theta){t_2}$$ let's try to get rid of the squared terms: $${t_1^2}={x_1-{V_0}\cos(\theta){t_1} \over A}\\ {t_2^2}={x_2-{V_0}\cos(\theta){t_2} \over A}\\ y_1=B{x_1-{V_0}\cos(\theta){t_1} \over A}+{V_0}\sin(\theta){t_1}\\ y_2=B{x_2-{V_0}\cos(\theta){t_2} \over A}+{V_0}\sin(\theta){t_2}$$ Now: $$y_1-x_1\frac{B}{A}={V_0}{t_1}\left(\sin(\theta)-{B \over A}\cos(\theta)\right)\\ y_2-x_2\frac{B}{A}={V_0}{t_2}\left(\sin(\theta)-{B \over A}\cos(\theta)\right)$$ so $$\frac{t_1}{t_2}=\frac{Ay_1-Bx_1}{Ay_2-Bx_2}=Q$$ Using the hint by @polfosol we can write: $$y_1-Bt_1^2=\tan(\theta)(x_1-At_1^2)\\ y_2-Bt_2^2=\tan(\theta)(x_2-At_2^2)$$ and rearranging $$y_1-x_1\tan(\theta)=(B-A\tan(\theta))t_1^2\\ y_2-x_2\tan(\theta)=(B-A\tan(\theta))Q^2t_1^2$$ from which you can easily isolate $$\tan(\theta)=\frac{Q^2y_1-y_2}{Q^2x_1-x_2}$$ From here is easy to find $t_1$, $t_2$, $\theta$ and, finally, $V_0$.