Transformed Parabola

Geometry Level pending

A parabola in the $xy$ plane is given by $v = \dfrac{1}{4} u^2$ , where the coordinates $(x, y)$ are related to the $(u,v)$ coordinates by

$\begin{bmatrix} x \\ y \end{bmatrix} = R \begin{bmatrix} u \\ v \end{bmatrix} + \begin{bmatrix} 5 \\ 2 \end{bmatrix}$

and $R$ is the $2 \times 2$ rotation matrix by $30^\circ$ counter clockwise, i.e.

$R =\begin{bmatrix} \cos 30^\circ && -\sin 30^\circ \\ \sin 30^\circ && \cos 30^\circ \end{bmatrix}$

Now, you apply the following transformation to the points on the parabola.

If $p$ is a point on the parabola, you define its image $p'$ as follows:

$p' = A p + b$

where

$A = \begin{bmatrix} 1 && -3 \\ 0 && 2 \end{bmatrix} , \hspace{24pt} b = \begin{bmatrix} 1 \\ - 4 \end{bmatrix}$

Given that the image of the parabola is another parabola, find the vertex of the new parabola and the direction of its axis. Let $r_0 = (r_{0x}, r_{0y} )$ be the vertex and $u = ( \cos \theta , \sin \theta )$ ( $\theta$ in radians, $0 \le \theta \lt 2 \pi$ ) be the unit vector along the axis pointing in the direction that the parabola opens up. Further, and just as important, by attaching a suitable reference coordinate frame $O' x' y'$ to the new parabola, where $O'$ is at the vertex, its equation can be written as

$y' = a {x'}^2$

with $a \gt 0$

Find $4(r_{0x} - r_{0y}) + \theta + a$

The answer is 7.5986.

1 solution

Hosam Hajjir
May 2, 2021

Here's an outline of one possible solution method that depends on parametric equations.

The coordinate vector in the $uv$ reference frame is given parametrically by $(t, \dfrac{1}{4} t^2) = t (1, 0) + t^2 (0, \dfrac{1}{4})$

Hence the coordinate vector in the $xy$ plane is given by

$p = (x, y) = R (u,v) + (5, 2) = R ( t (1, 0) + t^2 (0, \dfrac{1}{4} ) )+ (5, 2)$

Note that ordered pairs are actually $2 \times 1$ column vectors. Next apply the transformation to obtain

$p' = A p + b = A ( t R (1, 0) + t^2 R (0, \dfrac{1}{4} ) + (5, 2) ) + b$

and this can be written as

$p' = v_0 + v_1 t + v_2 t^2 \hspace{12pt} (1)$ ,

with $v_0 = A (5, 2) + b , v_1 = A R (1, 0) , v_2 = A R (0, \dfrac{1}{4} )$

Equation $(1)$ can be written as

$p' = u_0 + c_1 u_1 (t + t_0) + c_2 u_2 (t + t_0)^2 \hspace{12pt} (2)$

where $u_1$ and $u_2$ are unit vectors and mutually perpendicular to each other.

Expanding $(2)$ and equating similar terms, results in,

$v_0 = u_0 + c_1 u_1 t_0 + c_2 u_2 t_0^2 \hspace{16pt} (3)$

$v_1 = c_1 u_1 + 2 c_2 u_2 t_0 \hspace{16pt} (4)$

$v_2 = c_2 u_2 \hspace{16pt} (5)$

Equation $(5)$ determines $u_2$ and $c_2$ , as $c_2 = | v_2 |$ , and $u_2 = v_2 / c_2$

This also determines $u_1$ as a unit vector that is perpendicular to $u_2$ .

Next, from equation $(4)$ , by dot multiplying both sides with $u_1$ , we obtain

$v_1 \cdot u_1 = c_1$

and by dot multiplying both sides with $u_2$ , we obtain

$v_1 \cdot u_2 = 2 c_2 t_0$

hence $c_1$ and $t_0$ are now determined. Finally $u_0$ can be computed from equation $(3)$ .

The vertex is $u_0$ , $\theta$ is the angle of $u_2$ , and $a = \dfrac{c_2}{{c_1}^2}$

Performing these calculations gives us $r_0 = u_0 = (0.105328239,-0.437689286 )$ , $u_2 = (-0.872850734, 0.487987291)$ , and $a = 2.794707525$

Hence, $\theta = 2.631810294$ , and $4(r_{0x} - r_{0y} ) + \theta + a = 7.598587919 \approx \boxed{7.5986 }$

Transformed Parabola

The answer is 7.5986.

1 solution

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