Eccentricity of Pluto With an eccentricity of $0.25$, pluto's orbit is the most eccentric in the solar system. The length of the minor axis of its orbit is approximately $1×10^{10}$. Find the distance between Pluto and the sun at perihelion and at the aphelion {placing the origin at the centre of the orbit with the sun on the x axis}
Here is my attempt
Eccentricity $e = \frac{c}{a}\\$
Length of minor axis =$~1×10^9\\b=5×10^{10}\\c^2=a^2-b^2\\0.25=\frac{\pm\sqrt{a^2-b^2}}{a}\\a^2=\frac{(5×10^9)^2}{0.9375}\\a=\pm 5.164×10^9\\c=1.29×10^9$ $\pm5.164×10^9$ So my calculated distances between the sun and  Pluto at the aphelion  is  $6.454×10^9$ and $ 3.874×10^9$  at the  perihelion
How do I fix the errors??
 A: 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.
