Parabola (1 Viewer)

Paradoxica · Jul 13, 2016

juantheron said:
$The length of latus rectum of parabola whose parametric coordinates$

$are $x=at^2+bt+c$ and $y=a't^2+b't+c'$

The parametric equations are not in one of the two standard axial forms.

In order to solve this, we will reduce it to one of two such forms.

$\noindent Rotate the locus by $\theta : \\ \begin{bmatrix} x' \\ y' \end{bmatrix} \equiv \begin{bmatrix} x \\ y \end{bmatrix} \begin{bmatrix} \cos{\theta} & -\sin{\theta}\\ \cos{\theta} & \sin{\theta} \end{bmatrix} \\ $ Expand out the matrix to yield the following: $\\ x' = (a\cos{\theta} - a' \sin{\theta})t^2 + (b \cos{\theta} - b' \sin{\theta})t + c \cos{\theta} - c' \sin{\theta} \\ y' = (a' \cos{\theta} + a \sin{\theta})t^2 + (b' \cos{\theta} + b \sin{\theta})t + c' \cos{\theta} + c \sin{\theta} \\ $We may select either one to use as the linear parameter, so I will choose $x'. \\$So we set the co-efficient of $t^2$ in $x'$ to be zero and proceed to find everything we need that is relevant to $\theta. \\\\ a\cos{\theta} = a' \sin{\theta} \Rightarrow \tan{\theta} = \frac{a}{a'} \\ $The values of $\cos{\theta}, \sin{\theta}, \tan{\theta}$ are based upon the coefficients of the parametric equations, so the issue of signs and quadrants can be ignored.$

$$\noindent For brevity, we will use $\Delta^2$ in place of $a^2 + a'^2. \\\sin{\theta} = \frac{a}{\Delta}, \cos{\theta} = \frac{a'}{\Delta}, \tan{\theta} = \frac{a}{a'} \\ x' = \frac{a'b-ab'}{\Delta}t + \frac{a'c - ac'}{\Delta} \\ y' = \Delta t^2 + \frac{ab + a'b'}{\Delta}t + \frac{ac + a'c'}{\Delta} \\ $Complete the square on $y'$ in order to find the parameter $t'$ of the standard form: $x - x_0= 2at, y - y_0 = at^2 \\$On the other hand, notice that translation does not affect the geometric properties of the parabola, just like how rotation does not. So we can just complete the square and translate simultaneously by omitting everything except the leading terms on both parametric equations. Hence: $ \\ x' = \frac{a'b-ab'}{\Delta}t', \quad y' = \Delta t'^2 \\ $Eliminate the parameter $t'$ to bring it into Cartesian form.$ \\ \frac{y'}{\Delta} = \frac{\Delta^2}{(a'b-ab')^2}x^2 \Leftrightarrow x^2 = \frac{(a'b - ab')^2}{(a^2+a'^2)\sqrt{a^2+a'^2}}y$

$$\noindent Recall that in standard form, the Latus Rectum is the coefficient of $y$. Hence, $L = \frac{(a'b - ab')^2}{(a^2+a'^2)\sqrt{a^2+a'^2}}$

Paradoxica · Jul 13, 2016

I may have made mistakes. Feel free to point them out if you spot them.

Parabola (1 Viewer)

juantheron

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