Of the following vectors, parallel to the vector a = (2, − 3) is () A. （-4，6）B. （4，6）C. （-3，2）D. （3，2）

Of the following vectors, parallel to the vector a = (2, − 3) is () A. （-4，6）B. （4，6）C. （-3，2）D. （3，2）

For a, ∵ 2 × 6 = - 3 × (- 4) for B, ∵ 2 × 6 ≠ - 3 × 4, so it is not parallel. For C, ∵ 2 × 2 ≠ - 3 × (- 3), ∵ not parallel. For D, ∵ 2 × 2 ≠ - 3 × 3, ∵ not parallel, so (- 4,6) and a = (2, − 3), so select a

It is proved that (B, C, d) a + (C, a, d) B + (a, B, d) C + (B, a, c) d = 0, a, B, C, D are all vectors > 0

When you say (B, C, d) is a mixed product, that is to do cross product first and then dot product. The property of mixed product is: three vectors rotate in order, and the mixed product remains unchanged. For example, (B, C, d) = (D, B, C) and two vectors exchange in order, and the mixed product changes sign, for example, (B, C, d) = - (C, B, d) so (B, C, d) a + (C, a, d) B + (a, B, d) C + (b, a, c) d = (b

Quadratic equation x ^ 2 + (K + 9) x + 2K + 6 = 0, the sum of two squares is 24, find the value of K

x1+x2=-(k+9)
x1*x2=2k+6
x1^2+x2^2=(x1+x2)^2-2*x1*x2=(k+9)^2-2(2k+6)=(k+7)^2+20=24
(k+7)^2=4
K = - 5 or - 9

It is known that the two roots of the quadratic equation (6-k) (9-k) X2 - (117-15k) x + 54 = 0 with respect to X are integers, and the values of all real numbers K satisfying the conditions are obtained

The original equation can be reduced to: [(6-k) X-9] [(9-k) X-6] = 0. Because this equation is a quadratic equation of one variable with respect to x, K ≠ 6, K ≠ 9, then there are: X1 = 96 − K (1), X2 = 69 − K (2)

For the quadratic equation of one variable x2 + MX + n = 0, if two numbers are opposite to each other, then M=______ If two are reciprocal, then n=______ ．

Let two of the equations be X1 and X2 respectively, then: X1 + x2 = - M = 0, ｜ M = 0. X1 · x2 = n = 1, ｜ n = 1. So the answer is: M = 0, n = 1

Let a (m, - M + 3), B (2, m-1), C (- 1,4), the slope of the straight line AC = 3 times of the slope of the straight line BC, and find the value of the real number M

(-m+3-4)/(m+1)=3(m-1-4)/(2+1)
It is reduced to m ^ 2-3m-4 = 0
So m = 4 or - 1

A straight line with slope 1 passing through the left vertex a of the ellipse x2a2 + y2b2 = 1 (a > b > 0) has another intersection point m with the ellipse and B with the Y axis. If am = MB, the eccentricity of the ellipse is______ ．

According to the meaning of the question: left vertex a (- A, 0), the equation of line L is: y = x + a  B (0. A), and ∵ am = MB  m (− A2, A2) and ∵ m on the ellipse ∵ A2) 2A2 + (A2) 2B2 = 1, the answer is: A2 = 3B2 = 3 (a2-c2) ∵ 2A2 = 3c2  e = 63

(hope someone to answer, urgent, add) through the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) left vertex a slope of K straight line intersection ellipse
When passing through the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), the intersection ellipse C with the slope k of the left vertex A is at another point B, and the projection of point B on the X axis is just the right focus F, if 1 / 3 < K

Firstly, the coordinates of the intersection of the line and the ellipse are obtained, which are (- A, 0), (C, B ^ 2 / a) respectively;
Secondly, according to the slope formula k = (y1-y2) / (x1-x2), after applying the formula, we get that:
1/3

It is known that a vertex of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is a (2,0) and the eccentricity is √ 2 / 2. The line y = K (x-1) intersects ellipse C in two different directions
Point m, N, where the equation of ellipse C obtained from the first question is: x ^ 2 / 4 + y ^ 2 / 2 = 1
Find the value of K when the area of △ amn is √ 10 / 3

Let m (x1, Y1), n (X2, Y2)
When the ellipses are parallel, the following results are obtained
(1+2k²)x² - 4k²x+2k²-4=0
x1+x2=4k²/(1+2k²),x1x2=(2k²-4)/(1+2k²)
|MN|=√[(x1-x2)²+(y1-y2)²]
=√{ (x1-x2)² + [k(x1-1) - k(x2-1)]² }
=√[(x1-x2)² + k²(x1-x2)²]
=√[(1+k²)(x1-x2)²]
=√{ (1+k²)[(x1+x2)² - 4x1x2]
=√{ (1+k²)[16k^4/(1+2k²)² - 4(2k²-4)/(1+2k²) ] }
=√[(1+k²)(24k²+16)/(1+2k²)² ]
The distance from point a to the straight line is
h=|k|/√(1+k²)
∴S=(1/2)·h·|MN|
=(1/2)·[|k|/√(1+k²)] ·√[(1+k²)(24k²+16)/(1+2k²)² ]
=(1/2)·|k|·√[(24k²+16)/(1+2k²)²]
=√10/3
That is: | K · √ [(24K & # 178; + 16) / (1 + 2K & # 178;) &# 178;] = 2 √ 10 / 3
The square of both sides is: (24K ^ 4 + 16K & # 178;) / (1 + 2K & # 178;) - # 178; = 40 / 9
That is: 7K ^ 4 - 2K & # 178; - 5 = 0
The solution is: K & # 178; = 1 or - 5 / 7 (rounding off)
∴k²=1
∴k=±1

As shown in the figure, am is the tangent of ⊙ o, a is the tangent point, BD ⊥ am is at point D, BD intersects ⊙ o at point C, OC bisects ⊙ AOB, and calculates the degree of ⊙ B
Please write the detailed steps, thank you!

That's a simple question
OC is the angular bisector, so AOC = cob, BD is perpendicular to am, OA is perpendicular to am, so OA is parallel to BD, so AOC = OCB
OC = ob, so OCB = OBC is an equilateral triangle, 60 degrees