Hello, rednest!
The point $\displaystyle P(7\cos\theta,5\sin\theta)$ is on the ellipse with equation: $\displaystyle \frac{x^{2}}{49} + \frac{y^{2}}{25} = 1$
The line through $\displaystyle P$ parallel to the $\displaystyle y$axis meets the $\displaystyle x$axis at $\displaystyle X$
The point $\displaystyle Q$ is on the line $\displaystyle XP$ produced so that $\displaystyle XQ = 2XP$
Find, in cartesian form, an equation of the locus of $\displaystyle Q$, as $\displaystyle \theta$ 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: .$\displaystyle \begin{Bmatrix}x &=& 7\cos\theta \\ y &=&\text{}10\sin\theta \end{Bmatrix}$
Eliminate the parameter: . $\displaystyle \begin{Bmatrix}\dfrac{x}{7} &=&\cos\theta \\ \\[1mm]\text{}\dfrac{y}{10} &=&\sin\theta \end{Bmatrix}$
. . . . . . . . . . . . Square: . $\displaystyle \begin{Bmatrix}\dfrac{x^2}{49} &=& \cos^2\!\theta \\ \\[1mm] \dfrac{y^2}{100} &=&\sin\theta \end{Bmatrix}$
$\displaystyle \text{Add: }\;\;\frac{x^2}{49} + \frac{y^2}{100} \;=\;\underbrace{\cos^2\!\theta + \sin^2\!\theta}_{\text{This is 1}} $
Therefore: .$\displaystyle \boxed{\frac{x^2}{49} + \frac{y^2}{100} \;=\;1}$