It is known that the vertex of the parabola is at the origin and the focus is the focus of the hyperbola x ^ 2 / 16-y ^ 2 / 9 = 1

It is known that the vertex of the parabola is at the origin and the focus is the focus of the hyperbola x ^ 2 / 16-y ^ 2 / 9 = 1

Hyperbola x ^ 2 / 16-y ^ 2 / 9 = 1A & # 178; + B & # 178; = C & # 178; = 16 + 9 = 25, C = 5, so the focus coordinate (- 5,0) (5,0)
If you have not missed the title or copy wrong, then the following two cases
① When the focus coordinate (- 5,0) is Y & # 178; = - 2px, the focus coordinate (- P / 2,0) P = - 5 × - 2 = 10, Y & # 178; = - 20x
② When the focus coordinate (- 5,0) is Y & # 178; = 2px, the focus coordinate (P / 2,0) P = 5 × 2 = 10 y & # 178; = 20x

In the plane rectangular coordinate system, O is the coordinate origin, a, B, C three points satisfy the vector OC = 1 / 3oa + 2 / 3ob (all vectors). Prove that a, B, C three points are collinear

1. It is proved that OC = 1 / 3oa + 2 / 3ob can be changed into OC OA = 2 / 3 (OB OA), that is, AC = 2 / 3AB, which indicates that AC and ab are in the same direction, so ABC three points are collinear
2.oa=（1,COSX）,ob=（1+SINX,COSX）
Using the proof result AC = 2 / 3AB in (1), we can know that
oc=（SINX*2/3+1,COSX）
It can be seen that | ab | = √ (the square of SiNx) = SiNx (SiNx > 0 can be obtained from the condition that x belongs to [0, PI / 2]),
Then substitute the values of OA, OC and | ab | into f (x) = OA * oc - (2m ^ 2 + 2 / 3) * | ab |,
It is concluded that f (x) = 2 - (SiNx) ^ 2-sinx * 2m ^ 2,
Let SiNx = t, from 0 ≤ x ≤ π get 0 ≤ SiNx ≤ 1, that is, 0 ≤ t ≤ 1,
Then the original function can be changed to f (T) = - T ^ 2-2m ^ 2 * t + 2,
According to the properties of quadratic function, its symmetry axis t

Given that four points o, a, B and C in the plane satisfy vector 2oa + OC = 3ob, then the modulus of BC vector / the modulus of AB vector should be a number

It can be seen from the above that 2oa vector + OC vector = 3ob vector can be written as 2oa vector - 2ob vector + OC vector - ob vector = 0
Then we can get 2BA vector + BC vector = 0, so we can get the absolute value vector BC square / absolute value vector AB square = 4
Then we get the modulus of BC vector / the modulus of AB vector = 2, so this number is a number

Let o, a, B, C be four points in the same plane, vector OA = a, vector ob = B, vector OC = C, and a + B + C = 0, a * B + b * C + C * a = - 1, judge the shape of △ ABC
Pay attention to the symbols of the conditions given in the title!
In addition, if there is a process, thank you very much!

Let o, a, B, C be four points in the same plane, vector OA = a, vector ob = B, vector OC = C, and a + B + C = 0, a · B = B · C = C · a = - 1, judge the shape of triangle ABC
It's an isosceles triangle