Where is the vertex ?

\We have the following parametric equations:

$x(t) = x_0 + a t + b t^2$

$y(t) = y_0 + c t + d t^2$

We are going to create a new reference frame $x' y'$ that is rotated by an angle $\theta$ counter-clockwise, about the origin, with respect to the original reference frame. With this, we have

$x = c_1 x' - s_1 y'$

$y = s_1 x' + c_1 y'$

where $c_1 = \cos(\theta)$ and $s_1 = \sin(\theta)$

The above relations can be solved for $x'$ and $y'$ to yield,

$x' = c_1 x + s_1 y$

$y' = -s_1 x + c_1 y$

Using the expressions for $x(t)$ and $y(t)$ we obtain

$x' = c_1 (x_0 + at + b t^2 ) + s_1 ( y_0 + c t + d t^2 )$

$y' = -s_1(x_0 + at + bt^2) + c_1(y_0 + ct + dt^2 )$

We want to eliminate the coefficient of $t^2$ in $x'$ , so we have to choose $\theta$ such that

$b c_1 + d s_1 = 0$

that is,

$tan(\theta) = - b / d = - 3/2$

Therefore, $\theta = \tan^{-1}(-3/2) = -56.31^{\circ}$

With this choice, the expressions for $x'$ and $y'$ become,

$x' = (c_1 x_0 + s_1 y_0 ) + ( a c_1 + c s_1) t$

$y' = ( - s_1 x_0 + c_1 y_0 ) + (-s_1 a + c c_1) t + (-s_1 b + c_1 d ) t^2$

Note that $x'$ is now linearly related to the parameter $t$ , hence, the parabola is now in the upright position with respect to the $x' y'$ frame, and the vertex occurs when

$\dfrac{dy'}{dx'} = 0$

But,

$\dfrac{dy'}{dx'} = \dfrac{ \left( \dfrac{dy'}{dt} \right) }{ \left( \dfrac{dx'}{dt}\right) }$

Now, since $\dfrac{dx'}{dt}$ is a non-zero constant, then the vertex occurs when $\dfrac{dy'}{dt} = 0$ , i.e., at the critical point of $y'$ , which is at

$t_0 = -\dfrac{B}{2A} = - \dfrac{(-s_1 a + c c_1)}{2 ( -s_1 b + c_1 d ) }$

Substituting the given values, we obtain $t_0 = -0.153846$ , from which, $x' = -1.96278 , y' = 4.6296$ . And finally, we find that

$x = c_1 x' - s_1 y'$

$y = s_1 x' + c_1 y'$

so that, $x = 2.763314$ , $y = 4.201183$ , making the answer $2.763314 + 4.201183 = \boxed{6.9645}$

The given equations can be expressed in Cartesian coordinates as: $(2x-3y)^2+17x-50y+113=0$

We know standard formula of parabola:
$Y^2=4aX\equiv \text{(equation of symmetry axis)}^2+\text{(equation of tangent at vertex)}=0$

Now, introducing the arbitrary constant $t$ , the equation of parabola can be written as follows:

$(2x-3y+t)^2+(17-4t)x+(-50+6t)y+113-t^2=0$

select the value of $t$ such that $2(17-4t)-3(-50+6t)=0 \implies t=\frac{92}{13}$

Now, the equation of symmetry axis of parabola: $2x-3y+t=0\implies 2x-3y+\frac{92}{13}=0$

the equation of tangent at the vertex of parabola: $(17-4t)x+(-50+6t)y+113-t^2=0\implies -1911x+1274y-10633=0$

Now, solving equations of symmetry axis & tangent at the vertex, the coordinates of vertex are $x=2.763314, y=4.201183$

hence, $x+y=2.763314+4.201183=6.964494$