conic sectionsgeometry
I'm struggling with this question:
The point $P$ lies on the ellipse $x^2+4y^2=1$ and $N$ is the foot of the perpendicular from $P$ to the line $x=2$. Find the equation of the locus of the mid-point of $PN$ as $P$ moves on the ellipse.
It seems that the locus should be a circle, but I'm not sure how to find the actual equation for it. If $F$ is a focus, I know that the defining characteristic of an ellipse is that $0 < \frac{PN}{PS} < 1$. How do I solve this problem?
I should probably use the fact that $r/d = e$, where $r$ is the distance from the focus to any point $M(x,y)$ of an ellipse. $d$ the distance from $M(x,y)$ to the directrix, and $e$ is the eccentricity.
If you showed your attempt to begin with, we might be able to be a bit more helpful; alas, since you haven't shown it, you'll have to content yourself with the following sketch of a solution.
You know the distance from an arbitrary point $(x,y)$ to the focus $(3,0)$:
$$f=\sqrt{(x-3)^2+(y-0)^2}$$
and you can use the formula for point-line distance (formula 11 here) to get the distance from $(x,y)$ to the line $x+y-1=0$:
$$d=\frac{x+y-1}{\sqrt{1^2+1^2}}$$
from which you use the definition for eccentricity, $\varepsilon=\dfrac{f}{d}$, where $\varepsilon=\dfrac12$.
At once you should obtain an equation with a square root. You can try squaring both sides of the equation and then rearrange things to obtain a two-variable quadratic as usual, but you'll have to justify why the squaring is legal. You should end up with
$$7x^2-2xy+7y^2-46x+2y+71=0$$
I think it is a circle whose radius is $a$ and center is the origin.
As is shown in the graph, suppose C is a point on the ellipse, and F1,F2 are two focus, CM is the angle bisector of $\angle F_1CF_2$, and F2P is perpendicular to the tangent line, E lies at both line F1C and F2P.
According to the reflexive property, it is easy to know that CF2 = CE, and hence $\triangle CF_2P$ is congruent with $\triangle CEP$, so $P$ is the mid point of $F_2E$ and $O$ is the midpoint of $F_1F_2$, so $OP$ is always half the length of $F_1E$, and we know that $ CF_1 + CE = CF_1 + CF_2 = 2a$, so $OP = a$.
Best Answer
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.