Locus in the case of an ellipse.

Question

Given that $ S $ is the focus on the positive x-axis of the equation$ \frac{x^{2}}{25} + \frac{y^{2}}{9} =1$. Let $P=(5 \cos{t}, 3 \sin {t})$ on the ellipse, $SP$ is produced to $Q$ so that $PQ = 2PS$. Find the locus of $ Q $ if $P$ moves on the ellipse. I found that $S=(4,0)$ and try to use the line segment formula but cannot get the answer as stated below. The answer back is $\frac{(x+8)^{2}}{15} +\frac {y^{2}}{9}=1$ Can somebody help me? Thanks.

Answer 1

Homothety is a key word. You get a point $Q$ if you act on $P$ with a homothety with center at $S$ and factor $k=3$. So since homothety preserves shape, the result is again ellipse with a center at $(-8,0)$ and $a'=3a = 15$ and $b'=3b =9$. So we get $$ \frac{(x+8)^{2}}{15^2} +\frac {y^{2}}{9^2}=1$$

Answer 2

As $Q=P+2(P-S)$, the coordinates of $Q$ can be readily found to be: $$ x_Q=3x_P-8=15\cos t-8,\quad y_Q=3y_P=9\sin t. $$ It follows that $(x_Q+8)^2/15^2+y_Q^2/9^2=1$.