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.
Locus in the case of an ellipse.
conic sectionsgeometric transformationgeometrylocus
Related Solutions
Required to prove that the acute angles made by each foci to the tangent are equal
[Math] Find the equation of the locus of the mid-point between an elliptical point and its directrix
Hint: The line x=2 is a vertical line. If P=(x,y) is a point on the ellipse, the perpendicular to the line x=2 will always meet it at (2,y). So the mid-point is $(\frac{1}{2}(x+2), y)$.
Updated: If you're still stuck, here's the next step: use the parametrization of the ellipse $x=\cos(\theta)$, $y=\frac{1}{2}\sin(\theta)$ in the mid-point expression above. So the mid-point is $$ 1+\frac{\cos(\theta)}{2}, \frac{1}{2}\sin(\theta) $$
Now you can use $\cos^2 \theta + \sin^2 \theta= 1$ to get the equation of the locus.
Best Answer
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 $$ \frac{(x+8)^{2}}{15^2} +\frac {y^{2}}{9^2}=1$$