The vertices of a triangle are $(5, 3)$ and $(5, -5)$. Determine the equation of the locus of the third vertex if the perimeter of the triangle is $20$ units.
So I assume $5$ units to be the major segment where $b$ is the minor segment, and $3$ units is the distance to foci.
By Pythagoras, I got:
$a^2=b^2+c^2$
$b= 4$ units
But still, I can't get the right locus of points given with the choices. So I was thinking if the locus of points escribing or inscribing the triangle?
On
Your figure is not correct because the points $(5,3)$ and $(5,-5)$ are the foci of the ellipse. What you have is the center of the ellipse given by the midpoint $(5,-1)$. This said you that the ellipse is "vertical" instead of "horizontal".
You have the equation which is the definition itself of the ellipse as a locus $$\sqrt{(x-5)^2+(y-3)^2}+\sqrt{(x-5)^2+(y+5)^2}=20-8=12$$ Simplifying you get $$9(x-5)^2+5(y+1)^2=180$$ An ellipse "vertical" as you can verify.
Hint. Let $A=(5, 3)$ and $B=(5, -5)$. Let $P=(x,y)$ be the third point. Now impose that the perimeter of the triangle is $20$: $$\mbox{dist}(A,P)+\mbox{dist}(B,P)+8=20.$$ Recall that an ellipse is a plane curve such that the sum of the distances to the two focal points is constant for every point on the curve.