If the point m (3,0), the ellipse x24 + y2 = 1 and the line y = K (x + 3) intersect at points a and B, then the perimeter of △ ABM is () A. 4B. 8C. 12D. 16

The line y = K (x + 3) passes through the fixed point n (− 3, 0). From the definition of ellipse, we know that m and N are the focus of ellipse: an + am = 2A = 4, BM + BN = 2A = 4. The perimeter of △ ABM is ab + BM + am = (an + BN) + BM + am = (an + AM) + (BN + BM) = 8, so we choose B

Ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, F1, F2 are the left and right focus, P point is in the first quadrant, triangle pof2 area is an equilateral triangle with root 3, find B ^ 2
Two roots, three?
Point P is on the ellipse

The length of an equilateral triangle is 2
PO=PF2=F2O=F1O=2=c
So angle pf1o = angle f1po = 30
So pf1f2 is the RT triangle
According to Pythagorean theorem, Pf1 = 2, root 3
2A = Pf1 + PF2 = 2 + 2 3
B ^ 2 = a ^ 2-C ^ 2 = 4 + 2 root 3-4 = 2 root 3

Given that the ratio of distance between point m and two fixed points o (0,0), a (3,0) is 1 / 2, find the trajectory equation of point M

Let m (x, y)
So: ∣ Mo ∣ = √ (X & sup2; + Y & sup2;); ∣ Ma ∣ = √ [(x-3) & sup2; + Y & sup2;]
Because: ∣ Mo ∣ = 1 / 2 ∣ Ma ∣
So: √ (X & sup2; + Y & sup2;) = 1 / 2 √ [(x-3) & sup2; + Y & sup2;]
So: (x + 1) & sup2; + Y & sup2; = 4
That is: m trajectory is a circle with (- 1,0) as the center and 2 as the radius

Given that the distance ratio of m to two fixed points o (0,0) and a (3,0) is 1 / 2, the trajectory equation of M can be solved

Let m (x, y) | Mo | g | Ma = 1 / 2 | Mo | = √ x2 + Y2 | Ma | = √ (x-3) 2 + Y2
∴2√x2+y2=√(x-3)2+y2
It is reduced to 3x2 + 6x + 3y2-9 = 0

If the point m (3,0), the ellipse x24 + y2 = 1 and the line y = K (x + 3) intersect at points a and B, then the perimeter of △ ABM is () A. 4B. 8C. 12D. 16