If there is no real root in the quadratic equation x ^ 2 + (M + 2) x + (1 / 4) m ^ 2 = 0, the root number m ^ 2 + │ M-1 │

If there is no real root in the quadratic equation x ^ 2 + (M + 2) x + (1 / 4) m ^ 2 = 0, the root number m ^ 2 + │ M-1 │

If there is any problem, the general idea is as follows
Using the first conditional discriminant Δ = (M + 2) ^ 2-m ^ 2 = 4m + 4

(2006 Yantai) known: on the one variable quadratic equation X2 - (R + R) x + 1 4D2 = 0 has no real root, where R and R are the radii of ⊙ O1, ⊙ O2 respectively, and D is the center distance between the two circles, then the position relationship of ⊙ O1, ⊙ O2 is () A. Exotropism B. Tangency C. Intersection D. Contains

According to the meaning of the question, if the equation has no real root, we can get (R + R) 2-d2 < 0,
Then: (R + R + D) (R + R-D) < 0,
Because R + R + d > 0, R + R-D < 0,
That is, d > R + R,
Then, the two circles are separated
Therefore, a

It is known that the distance from the center of a circle with radius r to the line L is D, and the equation r square x + DX + 4 / 1 = 0 has two equal real roots Try to judge the position relationship between the line L and the center of the circle

Δ=d^2-R^2=0
D=R
Tangency

Given that the radius of ⊙ o is 1, the distance between point P and center O is m, and the equation x ⊙ 2x + M = 0 has two unequal real roots, try to determine the position of point P and ⊙ o

The equation has two unequal real roots. According to the following discriminant, 4-4m > 0, m < 1 is obtained. Therefore, the position relationship between point P and circle is that point P is in circle o,

We know that the absolute value of OA vector = 1, the absolute value of OB vector = root 3, the absolute value of OC vector = 1. The multiplication of OA vector by OB vector is equal to 0. Find the maximum value of Ca vector multiplied by CB vector

The maximum value of Ca vector multiplied by CB vector is 1 + radical 3

Given three points a (2,0), B (0,2), C (x, y), and the absolute value OA = 1, (1) if the absolute value vector OA + vector OC = radical 7 (o is Coordinate origin) to find the angle between vector OB and vector OC (2) If vector AC ⊥ vector BC, find the coordinates of point C

OA+OC=(2+x,y)
So x 2 + y 2 = 1, (2 + x) 2 + y 2 = 7
So x = 1 / 2, y = √ 3 / 2, so OC = (1 / 2, √ 3 / 2)
So ob * OC = 0 + √ 3 = √ 3, and | ob | = 2, | OC | = 1, so cos = √ 3 / 2, so the included angle is 30 degrees
（2）AC=(x-2,y) BC=(x,y-2)
Because the vector AC ⊥ vector BC, X (X-2) + y (Y-2) = 0, and X ⊥ y ⊥ is 1
So x = (1 + √ 7) / 4, y = (1 - √ 7) / 4 or x = (1 - √ 7) / 4, y = (1 + √ 7) / 4
So C ((1 + √ 7) / 4, (1 - √ 7) / 4) or C ((1 - √ 7) / 4, (1 + √ 7) / 4)

Given that the line ax + by + C = 0 and circle O, X2 + y2 = 4 intersect at two points of A. B and the absolute value AB = 2 times the root sign 3, calculate the values of vectors OA and ob

12＝|AB|²＝（OB-OA）²＝OB²+OA²-2[OA·OB]＝8-2[OA·OB],OA·OB＝-2

Given the vector OA ‖ ob, the absolute value vector OA = 3, the absolute value vector ob = 1, calculate the absolute value vector oa-ob

|Oa-ob | = 4 or 2

The midpoint of the quadrilateral ABCD is o known as OA = ob = OC = od = root 2 / 2. Is this quadrilateral a square?

Because the diagonals AC and BD of the quadrilateral ABCD intersect at point O, and OA = ob = OC = od = the root of the two halves AB, according to the square of the diagonal bisection is a parallelogram, the square of OA + the square of OB = the square of AB and the inverse Pythagorean theorem, the angle AOB = 90 degrees, so the quadrilateral ABCD with vertical and equal diagonals is a square

A focal point of the ellipse on the x-axis is perpendicular to the two ends of the minor axis, and the distance between the focus and the end point of the long axis is (root 10-radical 5). Find the standard equation

As shown in the figure, ∵ f1b1 ⊥ f1b2, easy to get ⊿ of 1B1 is isosceles right angle ⊿ B = C, a = √ 2. C also known ︱ f1a1 | = a-c = √ 10 - √ 5,  2  C-C = √ 5 (√ 2-1), the solution is C = √ 5, a = √ 10, B = C = √ 5 9 the elliptic equation is x