algebra-precalculusanalytic geometryconic sections
There are two lines, parallel to the $x$-axis, which pass through the foci and intersect the ellipse at four points. How can I find the points of intersection?
Let the foci be $F_1$ and $F_2$ and let a point of intersection be $P$. The tangent to the ellipse is the external angle bisector of $\angle F_1PF_2$, and the tangent to the hyperbola is the internal angle bisector.
Proving these statements without calculus is onerous. Here's a sketch of a partial proof for the statement about ellipses, to give the main ideas:
Let $F_1,F_2,P$ be as above. Let $b$ be the external bisector of $\angle F_1PF_2$; we want to show that $b$ is tangent to the ellipse. By the definition you've been given, that means we want to show that $b$ intersects the ellipse only at $P$. Consider first a point $Q$ on the other side of $b$ from the foci; I claim that $Q$ is not on the ellipse. To show this, draw $F_2Q$, meeting $b$ at $R$; also reflect $F_2$ in $b$ to obtain its image $F_2'$. Note that $F_1PF_2'$ are collinear because $b$ is the external angle bisector. Then $$ F_1Q+QF_2 = F_1Q + QR + RF_2 = F_1Q + QR + RF_2' > F_1F_2' = F_1P + PF_2' = F_1P + PF_2 $$ So $Q$ is not on the ellipse. I'm skipping some details here, notably why it has to be $>$ and not $\ge$ in the triangle inequality here. After you fill in that detail, and if you believe that the ellipse is a continuous curve, then we know that the ellipse doesn't cross $b$, so it is tangent to the ellipse. Then give a similar analysis for hyperbolas.
(We have not proved that conic sections have only one tangent at each point; I think I proved that once from your definition, but I only remember that it was unpleasant... and maybe my proof was wrong anyway.)
The nice way to prove these statements is with the notion of gradient from multivariable calculus: the direction of the gradient of the scalar field "distance from $F_i$" is obviously a unit vector pointing away from $F_i$; from this and the definition of an ellipse as a level curve of the sum of two such scalar fields it's clear that the normal to the ellipse bisects the desired angle. (On preview, along the lines of achille hui's answer.)
The co-ordinates of the focii are $(h\pm ae)$, so your $c$ will be $ae$ rather than $\frac{e}{a}$ and so your $a$ will be $\sqrt2$ and your $b$ will be $\sqrt\frac{3}{2}$. So the equation we have in the rotated system is:$$\frac{(x-\frac{1}{\sqrt2})^2}{2}+\frac{2y^2}{3}=1$$ Now all that remains is to somehow rotate this back into the original system.
Best Answer
You do not even need the equation for the ellipse if you use the property that, for any point $(x,y)$ on the ellipse, the sum of the distances from $(x,y)$ to the foci is a constant. Since we know that $(20\sqrt2,0)$ is a point on the ellipse, we can find this constant:
$\lVert (20\sqrt 2, -10)\rVert+\lVert(20\sqrt 2, 10)\rVert=2\sqrt{ 800+100}=60$.
Now we know that the y-coordinate of the points of intersection for the line going through $(0,10)$ is $10$, so we just need the x-coordinate. So we have that $$\lVert(x,10)-(0,10)\rVert + \lVert(x,10)-(0,-10)\rVert = 60\\ \lvert x \vert + \sqrt {x^2+400}=60\\ x^2+400=60^2-120\lvert x\rvert+x^2\\ \lvert x \rvert=\frac {3200}{120}=\frac {80} 3$$
So the points of intersection are $(\frac {80} 3 , 10)$ and $(\frac{-80} 3,10)$. The same method will allow you to find the other two points, try it! (You could also appeal to symmetry, but it's good to work through the argument on your own as well.)