Parametric and algebraic description of a parabola

The vertex of the parabola $\vec{r}\left( t \right)=\vec{C}+{{\vec{f}}_{1}}t+{{\vec{f}}_{2}}{{t}^{2}}$ is given by $\vec{r}\left( {{t}_{0}} \right)=\vec{C}+{{\vec{f}}_{1}}{{t}_{0}}+{{\vec{f}}_{2}}{{t}_{0}}^{2}$ , where ${{t}_{0}}=-\frac{{{{\vec{f}}}_{1}}\cdot {{{\vec{f}}}_{2}}}{2{{{\vec{f}}}_{2}}^{2}}$ (see here ).

Using $\vec{C}=\left( 5,7 \right)$ , ${{\vec{f}}_{1}}=\left( 3,1 \right)$ and ${{\vec{f}}_{2}}=\left( -1,2 \right)$ we get ${{t}_{0}}=\dfrac{1}{10}$ , thus, the vertex is $\vec{r}\left( {{t}_{0}} \right)=\left( 5.29,\ 7.12 \right)$ .

Furthermore, the focal length of the parabola is given by $f=\dfrac{{{{\vec{f}}}_{1}}^{2}{{{\vec{f}}}_{1}}^{2}-{{\left( {{{\vec{f}}}_{1}}\cdot {{{\vec{f}}}_{2}} \right)}^{2}}}{4{{\left| {{{\vec{f}}}_{2}} \right|}^{3}}}=\dfrac{49}{20\sqrt{5}}$ .

In the form ${y}'=a{{{x}'}^{2}}$ , $a=\dfrac{1}{4f}\Rightarrow a=\dfrac{5\sqrt{5}}{49}.$

Because ${{\vec{f}}_{2}}$ is parallel to the axis of the parabola, the angle ${{\theta }_{0}}$ is equal to the angle that ${{\vec{f}}_{2}}$ makes with $\vec{j}=\left( 0,1 \right)$ .

Hence, ${{\theta }_{0}}={{\cos }^{-1}}\left( \frac{{{{\vec{f}}}_{2}}\cdot \vec{j}}{\left| {\vec{f}} \right|\left| {\vec{j}} \right|} \right)={{\cos }^{-1}}\left( \frac{2}{\sqrt{5}} \right)$ .

For the answer, $a+{{x}_{0}}+{{y}_{0}}+{{\theta }_{0}}=\dfrac{5\sqrt{5}}{49}+5.29+7.12+{{\cos }^{-1}}\left( \frac{2}{\sqrt{5}} \right)\approx \boxed{13.1}.$

With $\mathbf{C} = (5, 7)$ , $\mathbf{f_1} = (3, 1)$ and $\mathbf{f_2} = (-1, 2)$ , we have parametric equations:

$x = 5 + 3t - t^2$

$y = 7 + t + 2t^2$

With a little bit of rearranging, we can eliminate $t$ and obtain:

$4x^2 + 4xy + y^2 - 61x - 55y + 401 = 0$

a conic where $A = 4$ , $B = 4$ , $C = 1$ , $D = -61$ , $E = -55$ , and $F = 401$ , and is rotated $\theta$ , where $\cot 2\theta = \frac{A - C}{B} = \frac{4 - 1}{4} = \frac{3}{4}$ . Using some trig identities, this means that:

$\sin \theta = \frac{\sqrt{5}}{5}$

$\cos \theta = \frac{2\sqrt{5}}{5}$

and applying the tranformation rotation :

$A' = A \cos^2 \theta + B \sin \theta \cos \theta + C \sin^2 \theta = 5$

$B' = 2(C - A) \sin \theta \cos \theta + B (\cos ^{2} \theta - \sin ^{2} \theta) = 0$

$C' = A \sin^2 \theta - B \sin \theta \cos \theta + C \cos^2 \theta = 0$

$D' = D \cos \theta + E \sin \theta = -\frac{177\sqrt{5}}{5}$

$E' = -D \sin \theta + E \cos \theta = -\frac{49\sqrt{5}}{5}$

$F' = F = 401$

so that a new graph with the same scale is:

$5x'^2 - \frac{177\sqrt{5}}{5}x' - \frac{49\sqrt{5}}{5}y' + 401 = 0$

which rearranges to:

$y' = \frac{5\sqrt{5}}{49}(x' - \frac{177\sqrt{5}}{50})^2 + \frac{179\sqrt{5}}{100}$

so that:

$a = \frac{5\sqrt{5}}{49}$

$x'_0 = \frac{177\sqrt{5}}{50}$

$y'_0 = \frac{179\sqrt{5}}{100}$

This also means that:

$x_0 = x'_0 \cos \theta - y'_0 \sin \theta = \frac{529}{100}$

$y_0 = x'_0 \sin \theta + y'_0 \cos \theta = \frac{178}{25}$

Therefore, $a = \frac{5\sqrt{5}}{49}$ , $x_0 = \frac{529}{100}$ , $y_0 = \frac{178}{25}$ , $\theta_0 = \theta = \cos^{-1} (\frac{2\sqrt{5}}{5})$ , and $a + x_0 + y_0 + \theta_0 = \frac{60809 + 500\sqrt{5}}{4900} + \cos^{-1} (\frac{2\sqrt{5}}{5}) \approx \boxed{13.1}$ .

Substituting $t = t' + d$ , then

$\mathbf{r}(t') = \mathbf{C}+ \mathbf{f_1}(t' + d) + \mathbf{f_2} (t'+d)^2$

$= \mathbf{V} + \mathbf{g_1} t' + \mathbf{g_2} t'^2$

where $\mathbf{V} = \mathbf{C} + \mathbf{f_1} d + \mathbf{f_2} d^2$ , $\mathbf{g_1} = \mathbf{f_1} + 2 d \mathbf{f_2}$ and $\mathbf{g_2} = \mathbf{f_2}$

We want $\mathbf{g_1}$ and $\mathbf{g_2}$ to be perpendicular to each other, so that $\mathbf{r}$ can be expressed in terms of two mutually perpendicular vectors comprising a (rotated) coordinate frame.

Setting $\mathbf{g_1} \cdot \mathbf{g_2} = 0$ gives us $(\mathbf{f_1} + 2 d \mathbf{f_2}) \cdot \mathbf{f_2} = 0$ , from which $d = -\dfrac{ \mathbf{f_1} \cdot \mathbf{f_2} }{ 2 \mathbf{f_2} \cdot \mathbf{f_2} }$

so that $\mathbf{V}$ , $\mathbf{g_1}$ and $\mathbf{g_2}$ are now fully specified.

Now let, $\mathbf{u_1} = \dfrac{ \mathbf{g_1} } { | \mathbf{g_1} | } , \mathbf{u_2} = \dfrac{ \mathbf{g_2} } { | \mathbf{g_2} | }$ , then

$\mathbf{r}(t') = \mathbf{V} + | \mathbf{g_1} | t' \mathbf{u_1} + | \mathbf{g_2} | t'^2 \mathbf{u_2}$

Defining $\mathbf{r'} = \mathbf{r} - \mathbf{V}$ , $x' = | \mathbf{g_1} | t'$ , $y' =| \mathbf{g_2} | t'^2$ , then

$\mathbf{r'} = [\mathbf{u_1}, \mathbf{u_2} ] [x', y']^T$

Hence, $(x', y')$ are the coordinates of $r$ with respect to a frame having its origin at $\mathbf{V}$ , and whose $x'$ axis is along $\mathbf{u_1}$ and $y'$ axis is along $\mathbf{u_2}$ .

The relation between $y'$ and $x'$ can be found by eliminating the parameter $t'$ , and one finds that $y' = a x'^2$ , where $a = \dfrac{| \mathbf{g_2} |} { | \mathbf{g_1} |^2}$

The angle $\theta_0$ is just the angle between $\mathbf{u_1}$ and the x-axis , so $\theta_0 = \cos^{-1}( \mathbf{u_1}_x )$ . The only thing to watch out for here is the case when $\mathbf{u_2}$ is a rotation by $-90^{\circ}$ from $\mathbf{u_1}$ (not $+90^{\circ}$ as it should be for a right handed coordinate system). In this case, we reverse the direction of $\mathbf{u_1}$ , so that $\mathbf{u_1} = - \dfrac{ \mathbf{g_1}}{ | \mathbf{g_1} |}$ . From the symmetry of the parabola this does not change the value of $a$ , but it does change $\theta_0$ .

Using the values for the vectors as specified in the problem, one obtains $a = 0.228170202 , \mathbf{V} = (5.29,7.12) , \theta_0 = 0.463647609$ , and this makes the answer $\boxed{13.1}$