## Solve Lesson 4 Page 48 Math Learning Topic 10 – Kite>

Topic

For ellipse (E): $$\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1$$. Find the coordinates of the point $$M \in \left( E \right)$$ such that the length $${F_2}M$$ is maximum, given that $${F_2}$$ is a focal point with positive coordinates of (E)

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: $$M{F_1} = a + \frac{c}{a}x;M{F_2} = a – \frac{c}{a}x.$$

$$M{F_1}$$ has a minimum value of $$a – c$$ when $$x = – a$$ and a maximum value of $$a + c$$ when $$x = a)$$

$$M{F_2}$$ has a minimum value of $$a – c$$ when $$x = a$$ and a maximum value of $$a + c$$ when $$x = – a)$$

Detailed explanation

The ellipse has the equation $$\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1 \Rightarrow a = 5,b = 3$$

We have: $$c = \sqrt {{a^2} – {b^2}} = \sqrt {25 – 9} = 4$$

Calling the coordinates of $$M(x,y)$$, we have: $$M{F_2} = a – \frac{c}{a}x = 5 – \frac{4}{5}x$$

Since $$– 5 \le x \le 5$$ or $$– 5 \le – x \le 5$$ $$5 + \frac{4}{5}\left( { – 5} \right) \le 5 + \frac{4}{5}( – x) \le 5 + \frac{4}{5}.5$$$$\Leftrightarrow 5 – 4 \le M{F_2} \le 5 + 4 \Leftrightarrow 1 \le M{F_2} \le 9$$

$$\Rightarrow M{F_2} \le 9$$. The equal sign occurs when $$– x = 5$$

So the maximum length $${F_2}M$$ is 9 when $$M( – 5.0)$$