Question:

The point $\displaystyle P(7 cos t,5 sin t)$ 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 $\displaystyle t$ varies.

2. Originally Posted by rednest
Question:

The point $\displaystyle P(7 cos t,5 sin t)$ 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 $\displaystyle t$ varies.
$\displaystyle P(7cost,5sint)$
$\displaystyle X(7cost,0)$
$\displaystyle Q(7cost,y)$

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

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

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

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