First slide

Ellipse

Question

$If P = (x, y), F_{1} = (3, 0), F_{2} = (- 3, 0), and 16 x^{2} + 25 y^{2} = 400, then P F_{1} + P F_{2} equals$

Moderate

A

8
B

6
C

10
D

12

Solution

$The ellipse can be written as$
$\begin{array}{l} \frac{x^{2}}{25} + \frac{y^{2}}{16} = 1 \\ Here, a^{2} = 25, b^{2} = 16 . Now, \\ b^{2} = a^{2} (1 - e^{2}) \\ or \frac{16}{25} = 1 - e^{2} \\ or e^{2} = 1 - \frac{16}{25} = \frac{9}{25} \\ or e = \frac{3}{5} \end{array}$
$The foci of the ellipse are (\pm a e, 0) \equiv (\pm 3, 0), i.e., F_{1} and F_{2} are the foci of the ellipse.$
$Therefore, we have P F_{1} + P F_{2} = 2 a = 10 for every point P on the ellipse.$

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