The equation $$ax^2+2hxy+by^2=c$$ represents a conic rotated about the axis (depending on $a,b,h,c$) with centre $(0,0)$.
You can always rotate the axis to remove the $xy$ term. This makes it easy to find the focus, length of major axis, etc.
To remove $xy$ term without shifting origin, lets rotate the axis by an angle $\theta$.
Let $X,Y$ represent the new coordinate system. Then,
$$x=X\cos\theta-Y\sin\theta$$
$$y=X\sin\theta+Y\cos\theta$$
Substituting and letting coefficient of $XY=0$, you get $$\tan2\theta=\frac{2h}{a-b}$$
For example, lets find $e$ of the following equation (which represents an ellipse): $$4x^2+2y^2+2\sqrt{3}xy=1$$
Raotate the axis by $$\tan2\theta=\frac{2h}{a-b}=\sqrt{3}$$
$$\theta=\frac{\pi}{6}$$
Now, substitute $$x=X\frac{\sqrt{3}}{2}-Y\frac{1}{2}$$
$$y=Y\frac{\sqrt{3}}{2}+X\frac{1}{2}$$
You will get $$5X^2+Y^2=5$$
Now you can find eccentricity using the formula you specified. (Rotating the ellipse or hyperbola will not change the eccentricity and the length of major\minor axis.
This is the graph for the two equations:
(Black represents the new ellipse)
If I understood your question correctly, you're essentially asking how one can find the equation for the directrix if one only has the equation for an ellipse with a given eccentricity.
You start with the equation below
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\qquad(1)
$$
where $a$ and $b$ are positive real-valued constants.
If you then define two points on the $x$ axis, $F$ and $F'$, by their coordinates,
$$
F \equiv (\epsilon a, 0)
\qquad(2)
$$
$$
F' \equiv (-\epsilon a, 0)
\qquad(3)
$$
for some as-yet undetermined real value $\epsilon \ge 0$, and then require that the sum of the distances from $F$ to $P$ and from $P$ to $F'$ be equal to $2a$ - for any arbitrary point $P$ on the ellipse - you'll find out that that is only possible if you choose $\epsilon$ to satisfy
$$
\epsilon = \sqrt{1 - \frac{b^2}{a^2}}
\qquad(4)
$$
So far, the above has nothing to do with the directrix. It's just a way to construct the foci and find the eccentricity, starting from some equation and then requiring that the curve described by that equation must have the fundamental property of an ellipse (namely, that the sum of the distances from any point on the ellipse to the two foci is a constant).
Now, let's see how we can prove the directrix property. Imagine that there is a vertical line at $x=d$ for some as-yet-undetermined real value $d \ge 0$ and let's see if we can find a value of $d$ such that the directrix property is satisfied. The focus for positive $x$ is $F$ (see its coordinates above), and an arbitrary point $D$ in the would-be directrix has coordinates $D = (d, y)$. The directrix property mandates that
$$
\epsilon = \frac{\overline{PF}}{\overline{PD}}
\qquad(5)
$$
Now, let's look at the square of the distances involved, since that gets rid of the square roots:
$$
\overline{PF}^{\,2} = (x - \epsilon a)^2 + (y - 0)^2 = (x - \epsilon a)^2 + y^2
$$
But $P$ lies on the ellipse so $(x,y)$ satisfies $(1)$. Therefore, eliminating $x^2$ in favour of $y^2$, but leaving the $x$ term alone, we have
$$
\overline{PF}^{\,2} =
a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax
$$
How about $\overline{PD}^{\,2}$? Note that $D$ and $P$ have the same $y$ coordinate so
$$
\overline{PD}^{\,2} = (x - d)^2 + (0)^2 = (x - d\,)^2
$$
Here's the crux now. Can we find a value of $d$ such that $(5)$ is true?
Imposing
$$
\frac{\overline{PF}^{\,2}}{\overline{PD}^{\,2}} =
\frac{a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax}{(x - d\,)^2} = \epsilon^2
$$
we get
$$
a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax = \epsilon^2\,(x - d\,)^2
$$
Note that the complicated expression above is something like this:
$$
(\mbox{terms independent of $x$ and $y$}) + (\mbox{terms involving $y^2$}) - 2\epsilon ax + 2\epsilon^2 xd = 0
$$
because the $x^2$ terms can be removed using $(1)$. Since the above has to be true for all $x$, the only hope for $(5)$ to be possible is if we choose $d$ such that
$$
2\epsilon ax - 2\epsilon^2 xd = 0
\qquad\Rightarrow\qquad
d = \frac{a}{\epsilon}
$$
With this choice, it's then not hard to show that the terms not written above also vanish so, indeed,
$$
\epsilon = \frac{\overline{PF}}{\overline{PD}}
$$
is true, provided we make that choice for $d$.
So, what's the conclusion? The conclusion is that an ellipse has two equivalent properties, namely, (A) the sum of the distances from the foci to any arbitrary point on the ellipse is a constant, and (B) the ratio of (the distance from an arbitrary point on the ellipse to one of the foci) to (the distance from that point to a fixed vertical line parallel to the semi-minor axis) is equal to the eccentricity of the ellipse.
And we also proved that $\epsilon = a/d$.
(Technically, I only proved that (A) above implies (B). We'd also have to prove that (B) implies (A) for the equivalence I referred to be proven. That's an exercise I leave for the reader...)
Best Answer
Just as a common sense check, the distance to the sun at perihelion must be less than half the minor axis; it would be half the axis if the sun were at the center, and it's at a focus instead, closer to the ellipse.
I don't know what most of the letters you used refer to (some description or a picture would help here), so I'll define and use my own terminology to describe the key properties of the ellipse.
Let $D$ be the major axis, $d$ be the minor axis, and $f$ be the focal length - the distance between the two foci. The eccentricity $\epsilon$ is the ratio $\frac{f}{D}$. Perihelion and aphelion both come on the major axis, with distances $\frac{D-f}{2}=\frac{1-\epsilon}{2}D$ and $\frac{D+f}{2}=\frac{1+\epsilon}{2}D$. A quick check - you've got the ratio between these two right, at least.
Then, by the defining equal-distance property of the ellipse, the distance from a focus to an endpoint of the minor axis is $\frac D2$. That's the hypotenuse of a right triangle with legs $\frac f2$ and $\frac d2$, so $f^2+d^2=D^2$ and $d^2=(1-\epsilon^2)D^2$. Solving, $D=\frac1{\sqrt{1-\epsilon^2}}d$. In this case, $\epsilon^2\approx\frac1{16}$, so we multiply $d$ by $\sqrt{\frac{16}{15}}\approx 1.033$ to get $D\approx 1.033\cdot 10^9$. Multiply that by $\frac{1+0.25}{2}$ and $\frac{1-0.25}{2}$ for an aphelion distance of $0.645\cdot 10^9 = 6.45\cdot 10^8$ and a perihelion distance of $0.387\cdot 10^9 = 3.87\cdot 10^8$.
So then, where did you go wrong? These numbers should all look very familiar - you're off by exactly a factor of $10$. And that factor comes in your very first step, dividing by $2$ to get a semi-minor axis of $5\cdot 10^9$ instead of the correct $5\cdot 10^8$.
So, looking it up, I have to wonder what units this is supposed to be in. The eccentricity is close enough to accurate, but the actual major axis is about $11.8\cdot 10^9$ km. It looks like the writer of the problem just picked a convenient-looking number out of thin air.