If the sum of the distances from a point on the ellipse to two focal points (- 4,0), (4,0) is equal to 10, then the length of the minor axis of the ellipse is

If the sum of the distances from a point on the ellipse to two focal points (- 4,0), (4,0) is equal to 10, then the length of the minor axis of the ellipse is

2a=10；
a=5；
c=4；
So the short axis length = √ (5 & # 178; - 4 & # 178;) = 3;
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Given that P is the point on the ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1, F1F2 is the left and right focus, and the angle f1pf2 = 60, calculate the coordinates of P

S △ f1pf2 = B ^ 2 * Tan @ / 2 (focus triangle area formula)
So s △ f1pf2 = 3 times root 3
And because s △ f1pf2 = 1 / 2 * 2C * h, H = 3 times root 3 / 4
H is the ordinate of point P, which can be solved by the equation
P (- 15 / 4,3-fold root 3 / 4)
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Let p be a point on the square of the ellipse X / 4 + the square of y = 1, (x ^ 2 / 4 + y ^ 2 = 1), and F1 and F2 are the two focal points of the ellipse, then what is the minimum value of | Pf1 | PF2 |

The definition of ellipse is defined by the ellipse, where 124; Pf1 124124124; PF2 | PF2 | PF2 | PF2 | 2 | PF2 124124\124; PF2 | 2 | PF2 | = 2A = 4, and | f1f1f2 F1F2 124124124\\\\124\124\\\\\theminimum value of | PF2 | is 1

It is known that f 1.f 2 is the two focal points of the ellipse with X squared quarter and y = 1, and P is a moving point on the ellipse, then the maximum value of | Pf1 | multiplied by | PF2 | is

|PF1|+|PF2|=4
(|PF1|+|PF2|)^2=|PF1|^2+|PF2|^2+2|PF1||PF2|=16>4|PF1||PF2|
|PF1||PF2|