## Solve Lesson 3 Page 48 Math Learning Topic 10 – Kite>

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.