It is known that a vertex of an ellipse is a (0, - 1) and the focus is on the x-axis. If the distance between the right focus and the root of the line X-Y + 2 is 2 = 0, the distance is 3 Let the ellipse intersect with the straight line y = KX + m and have different two points m n when am = an to find the value range of M

It is known that a vertex of an ellipse is a (0, - 1) and the focus is on the x-axis. If the distance between the right focus and the root of the line X-Y + 2 is 2 = 0, the distance is 3 Let the ellipse intersect with the straight line y = KX + m and have different two points m n when am = an to find the value range of M

If the right focus is set to F2 (C, 0), then | C + 2 √ 2 | / √ 2 = 3, the solution is C = √ 2, and a (0, - 1) is the vertex, so B = 1, so a = √ 3, then the elliptic equation is: X ﹣ 3Y | 2 = 3 ① Let m (x1, Y1), n (X2, Y2) and the midpoint Q (x0, Y0) of Mn, then X1? 2 + 3Y1? 2 = 3 ②x2²+3y2&su...

The angle BOC is in the plane α, a is the oblique line of plane α, and the angle AOB = angle AOC = 60 °, OA = ob = OC = α, BC = radical 2 α Find the angle between OA and plane α!

Because ob = OC, so OD ⊥ BC. And because ∠ AOB = ∠ AOC = 60 °, so AB = AC = = > ad ⊥ BC = = > plane AOD ⊥ plane α OA = ob = a, ∠ AOB = 60 ° then the triangle AOB is equilateral triangle

It is known that OA ⊥ OC, and ∠ AOB: ∠ AOC = 2:3, then the degree of ∠ BOC is () A. 30° B. 150° C. 30 ° or 150 ' D. 90°

∵OA⊥OC，
∴∠AOC=90°，
∵∠AOB：∠AOC=2：3，
∴∠AOB=60°．
Because there are two kinds of position of ∠ AOB: one is inside ∠ AOC, the other is outside ∠ AOC
① When in ∠ AOC, ∠ BOC = 90 ° - 60 ° = 30 °;
② When it is outside of AOC, BOC = 90 ° + 60 ° = 150 °
Therefore, C

Given a ray OA, if two more rays ob are drawn from point O, and OC make the angle AOB = 60 ° and the angle BOC = 20 °, find the degree of the angle AOC Seek geometry language, seem to be two results!

You can draw the picture yourself
First, we determine that AOB = 60 ° and the two results are due to the position of the ray OC from point o
1. When the ray OC is between the ray OA and the ray ob, then the angle AOB = 60 ° and the angle BOC = 20 °, so the angle AOC = 60 ° - 20 ° = 40 °;
2. Ray OC is not between ray OA and ray ob, and the angle AOC is 60 ° + 20 ° = 80 °

If ∠ BOC is in plane α, OA is a diagonal line of plane α. If ∠ AOB = ∠ AOC = 60 °, OA = ob = ob = a, BC = = √ 2a, find the angle formed by OA and plane α

Take the midpoint D of BC to connect OD, AD.OA=OB=OB=a In this paper, we deduce that the triangle AOC and the triangle AOB are equilateral triangles, then AB = AC = a, BC = = √ 2A. It is concluded that the BOC and BAC of the triangle are equilateral right triangles. We can calculate the side length of the triangle and get ad = od = = = √ 2 / 2A. Because OA = a, we can conclude that the triangle ADO is a right angle equilateral triangle,
As a matter of fact, we can see from the picture that if ad is perpendicular to BC and OD, AOC is the angle formed by OA and plane α is 45 degrees

If the angle AOB = angle AOC = 60 degrees, OA = ob = OC = 1, BC = radical 2, then the degree of the angle formed by OA and plane t is?

From the angle AOB = angle AOC = 60 degrees, OA = ob = OC = 1, we can get that the triangle AOB and AOC are equilateral triangles, so AB = AC = 1, do the height on the bottom of two triangles BC, because they are isosceles triangles, so the height can meet a point on BC, set as D, it is easy to find that ad = od = root sign 1 / 2, and AO = 1, so ad is perpendicular to OD, and

If there are at least three different points on the circle x ^ 2 + y ^ 2-4x-4y-10 = 0 and the distance from the straight line L: ax + by = 0 is twice the root 2, then what is the value range of the slope of the straight line l? (negative infinity, 2-3 ^ 0.5] ∪ [2 + 3 ^ 0.5 positive infinity)

If the circle equation is (X-2) ^ 2 + (Y-2) ^ 2 = 18, the center of the circle is (2,2), and the radius is r = 3 * 2 ^ 0.5. The straight line equation passing through the origin can be directly written as: y = KX, that is, kx-y = 0, where k = - A / b. if there are at least three points on the circle whose distance from the straight line is d = 2 * 2 ^ 0.5, then the straight line must intersect the circle, otherwise, the point satisfying the condition

If there are at least three different points on the circle x ^ 2 + y ^ 2-4x-4y-10 = 0 and the distance from the line L: ax + by = 0 is two radical two, then

Then what?
Circle (X-2) 2 ＋ (Y-2) 2 = (3 √ 2) 2
Center (2,2), radius 3 √ 2
|2a＋2b|/√(a²＋b²)≤√2
∴(a/b)²＋4(a/b)＋1≤0
∴－2－√3≤a/b≤－2＋√3,k＝－a/b
∴2－√3≤k≤2＋√3
The range of inclination angle: [π / 12,5 π / 12]

If there are at least three different points on the circle x ^ 2 + y ^ 2-4x-4y-10 = 0 and the distance from the line L: ax + by = 0 is two root sign two, then the value range of the inclination angle of the straight line L is

This is the problem of finding the zero bound value. The requirement is that when only one point on one side of the line meets the requirements, the slope of the line can be obtained. The length of the chord is 8, and then the equation of the straight line and the circle is combined to obtain - A / B

If there are at least three different points on the square of the circle x plus the square of Y minus 4x minus 4Y minus 10 = 0, the distance from the straight line y = KX is 2 times the root sign, then the value range of K is

x²+y²-4x-4y-10=0
(x-2)² + (y-2)² = 18
Radius of center (2,2) 3 √ 2
Combination of number and shape
The critical case is that the distance from the center of the circle to y = KX is √ 2
Easy to get k = 2 ± √ 3
According to the graph, the value range of K satisfying the condition is [2 - √ 3,2 + √ 3]