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

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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 \(M( – 5.0)\)

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