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If a line passing through a focus F1 of ellipse 4x2 + 2Y2 = 1 intersects with ellipse at two points a and B, then a and B and another focus F2 of ellipse form △ abf2, then the perimeter of △ abf2 is () A. 2B. 22C. 2D. 1
Ellipse 4x2 + 2y22 = 1, that is & nbsp; x214 + y212 = & nbsp; 1, a = 22, B = 12, C = 12. The circumference of △ abf2 is (| AF1 | + | af2 | + (| BF1 | + | BF2 |) = 2A + 2A = 4A = 22, so B
If a line passing through a focus F1 of ellipse 4x2 + 2Y2 = 1 intersects with ellipse at two points a and B, then a and B and another focus F2 of ellipse form △ abf2, then the perimeter of △ abf2 is ()
A. 2B. 22C. 2D. 1
Ellipse 4x2 + 2y22 = 1, that is & nbsp; x214 + y212 = & nbsp; 1, a = 22, B = 12, C = 12. The circumference of △ abf2 is (| AF1 | + | af2 | + (| BF1 | + | BF2 |) = 2A + 2A = 4A = 22, so B
Given that the two focal points of ellipse C passing through point m (1,3 / 2) are a (- 1,0) B (1,0) o as the coordinate origin, the equation of ellipse C is solved
Urgent need
Points m and B are on the line x = 1
MB=3/2
2c=1+1=2
According to Pythagorean theorem
MA=√2²+（3/2）²=5/2
So 2A = ma + MB = 5 / 2 + 3 / 2 = 4
a=2
b²=a²-c²=2²-1²=3
Elliptic equation: X & sup2 / 4 + Y & sup2 / 3 = 1
The combination of number and shape will reduce your calculation
Given that the two focal points of ellipse C are (- 1,0) and (1,0) and pass through point a (1,3 / 2), O is the coordinate origin, the equation of ellipse C is solved
AF1 = radical (1 + 1) ^ 2 + (3 / 2) ^ 2 = radical (4 + 9 / 4) = 5 / 2
Af2 = root (1-1) ^ 2 + (3 / 2) ^ 2 = 3 / 2
So, AF1 + af2 = 2A = 5 / 2 + 3 / 2, a = 2
b^2=a^2-c^2=4-1=3
So the elliptic equation is x ^ 2 / 4 + y ^ 2 / 3 = 1
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