**Topic**

The Earth moves around the Sun in an elliptical orbit of which the Sun is the focal point. Given that this ellipse has a semi-major axis \(a \approx 149.598.261\) km and an eccentricity \(e \approx 0.017\). Find the minimum and maximum distances between the Earth and the Sun (results rounded to the nearest unit)

**Solution method – See details**

For ellipse(E): \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) \((0 < b < a)\)

+ The length of the radius through the focal point of the point \(M(x,y)\) on (E) is: \(M{F_1} = a + \frac{c}{a}x;M{F_2} = a – \frac{c}{a}x.\)

**Detailed explanation**

Choose the coordinate system so that the Sun coincides with the focal point \({F_1}\) of the ellipse. Then, applying the focal radius formula, we have the distance between the Earth and the Sun as:

\(M{F_1} = a + ex\) where \(x\) is the coordinate representing the Earth as \( – a \le x \le a\)

Since \( – a \le x \le a\) \(a + e\left( { – a} \right) \le a + \frac{c}{a}x \le a + e\left( a \right) \Leftrightarrow a – ae \le a + \frac{c}{a}x \le a + ae\)

With the semi-major axis \(a \approx 149\;598\;261\) km and the eccentricity \(e \approx 0.017\), we have:

\(147\;055\;090 \le M{F_1} \le 152\;141\;431\)

So the minimum and maximum distances between the Earth and the Sun are 147 055 090 km and 152 141 431 km, respectively.