Question:

The point $P(7 cos t,5 sin t)$ is on the ellipse with equation $\frac{x^{2}}{49} + \frac{y^{2}}{25} = 1$. The line through $P$ parallel to the $y$-axis meets the $x-axis$ at $X$. The point $Q$ is on the line $XP$ produced so that $XQ = 2XP$. Find, in cartesian form, an equation of the locus of $Q$, as $t$ varies.

2. Originally Posted by rednest
Question:

The point $P(7 cos t,5 sin t)$ is on the ellipse with equation $\frac{x^{2}}{49} + \frac{y^{2}}{25} = 1$. The line through $P$ parallel to the $y$-axis meets the $x-axis$ at $X$. The point $Q$ is on the line $XP$ produced so that $XQ = 2XP$. Find, in cartesian form, an equation of the locus of $Q$, as $t$ varies.
$P(7cost,5sint)$
$X(7cost,0)$
$Q(7cost,y)$

$\frac{x^{2}}{49} + \frac{y^{2}}{25} = 1$
$\frac{(7cost)^{2}}{49} + \frac{(5sint)^{2}}{25} = 1$
$cos^2t+sin^2t= 1$

$XQ=2XP$
$XQ^2=4XP^2$
$(7cost-7cost)^2+(y-0)^2=4[(7cost-7cost)^2+(5sint-0)^2]$
$y^2=100sin^2t$
$y=10sint$

What is cartesian form?
Is it $Q(7cost,10sint)$?

3. Hello, rednest!

The point $P(7\cos\theta,5\sin\theta)$ is on the ellipse with equation: $\frac{x^{2}}{49} + \frac{y^{2}}{25} = 1$

The line through $P$ parallel to the $y$-axis meets the $x$-axis at $X$

The point $Q$ is on the line $XP$ produced so that $XQ = 2XP$

Find, in cartesian form, an equation of the locus of $Q$, as $\theta$ varies.
Code:
                    |
* * *
*       |       *   P
*          |          o      -
*            |       *  | *    :
|     *    |      : y
*             |  * θ     |  *   :
----*-------------+----------+--*----
*             |         X|  *   :
|          |      :
*            |          | *    :
*          |          *      : 2y
*       |       *  |      :
* * *        |      :
|          |      :
|          |      :
|          o      -
|          Q

After all that sketching and a lot of thinking, it turns out that

. . the equations for Q are: . $\begin{Bmatrix}x &=& 7\cos\theta \\ y &=&\text{-}10\sin\theta \end{Bmatrix}$

Eliminate the parameter: . $\begin{Bmatrix}\dfrac{x}{7} &=&\cos\theta \\ \\[-1mm]\text{-}\dfrac{y}{10} &=&\sin\theta \end{Bmatrix}$

. . . . . . . . . . . . Square: . $\begin{Bmatrix}\dfrac{x^2}{49} &=& \cos^2\!\theta \\ \\[-1mm] \dfrac{y^2}{100} &=&\sin\theta \end{Bmatrix}$

$\text{Add: }\;\;\frac{x^2}{49} + \frac{y^2}{100} \;=\;\underbrace{\cos^2\!\theta + \sin^2\!\theta}_{\text{This is 1}}$

Therefore: . $\boxed{\frac{x^2}{49} + \frac{y^2}{100} \;=\;1}$