Let L: y = KX + 2 and the square of ellipse C: 2 / x plus the square of Y 1 intersect at two different points a, B, O as the origin of coordinates, (1) find the straight range of K; (2) if OA. Ob = 1, find the equation of line L; (3) when k is the value, the area of triangle OAB will get the maximum value;

Let L: y = KX + 2 and the square of ellipse C: 2 / x plus the square of Y 1 intersect at two different points a, B, O as the origin of coordinates, (1) find the straight range of K; (2) if OA. Ob = 1, find the equation of line L; (3) when k is the value, the area of triangle OAB will get the maximum value;

Let a (x1, Y1), B (X2, Y2), then X1 + x2 = - 8K / (2k ^ 2 + 1); (1) x1x2 = 6 / (2k ^ 2 + 1); (2) OA * ob = x1x2 + y1y2 = (k ^ 2 + 1) x1x2 + 2K (x1 + x2) + 4 = 1, (3) (1), (2) substitute

It is known that a straight line y = - x plus m intersects an ellipse x2 / 4 + Y2 / 2 = 1 at two points a and B, if AB is a circle of diameter passing through the origin
It is known that the line y = - x plus m intersects the ellipse x2 / 4 + Y2 / 2 = 1 at two points a and B. If AB is a circle with diameter passing through the origin, the value of M is obtained

x²/4+y²/2=1
Substituting y = - x + m to get
3x²-4mx+2m²-4=0
x1+x2=4m/3
x1x2=(2m²-4)/3
AB is the diameter of the circle passing through the origin
That is, vector OA * vector ob = 0
Let a (x1, Y1), B (X2, Y2)
x1x2+y1y2=0
x1x2+(-x1+m)(-x2+m)=0
2x1x2-m(x1+x2)+m²=0
2(2m²-4)/3-4m²/3+m²=0
The solution is m = ± 2 √ 6 / 3

If the line y = KX + 1 and ellipse x2 + 10y2 = 5 intersect at two points a and B, if the circle with diameter AB passes through the origin, the lab equation is solved
The equation of lab, the equation of line L

The distance from the coordinates of the midpoint of AB to the origin is equal to half of AB (the circle passes through the origin, so ABO is a right triangle). The linear equation is substituted into the elliptic equation. The distance from AB and the distance from the midpoint of AB to the origin can be expressed by x1x2. Finally, an equation about K is solved and then replaced