Given the circle C: (X-30) ^ 2 + y ^ 2 = 100 and the point a (- 3,0), P is any point on the circle C, the vertical bisector l of the line PA intersects with PC at point Q, and the trajectory equation of point q is obtained It's x-3, not X-30

Given the circle C: (X-30) ^ 2 + y ^ 2 = 100 and the point a (- 3,0), P is any point on the circle C, the vertical bisector l of the line PA intersects with PC at point Q, and the trajectory equation of point q is obtained It's x-3, not X-30

Circle C Center (3,0) radius 10
As shown in the figure below, PQ = QA
Then QA + QC = QP + QC = 10
The locus of Q point is an ellipse with focal length of 2 * 3 and long axis of 10
Elliptic equation x / 25 + Y / 16 = 1

Hyperbola known fixed point P (- 4,0) and fixed circle Q: X & # 178; + Y & # 178; = 8x, moving circle m and circle Q tangent, and through the point P, find the center of the circle M
Given the fixed point P (- 4,0) and the fixed circle Q: X & # 178; + Y & # 178; = 8x, the moving circle m is tangent to the circle Q, and through the point P, the trajectory equation of the center m is obtained

The difference between the distance from point m to point P and the distance from point m to point q is constant R = 4
Then the trajectory of the moving point m is a hyperbola with focus on M and Q and 2A = 4
The results are as follows
a=2、c=4
The trajectory equation of the obtained point is as follows:
x²/4－y²/12=1

Given that the moving circle m is tangent to the straight line y = 3 and circumscribed to the fixed circle C: x2 + (y + 3) 2 = 1, the trajectory equation of the moving circle center m is obtained

Let the center of the moving circle be m (x, y) and the radius be r, then the distance from m to C (0, - 3) is equal to the distance to the straight line y = 3. (4 points) according to the definition of parabola, the trajectory of the center of the moving circle is a parabola with C (0, - 3) as the focus and y = 3 as the guide line. (8 points) its equation is x2 = - 12Y. (12 points)

If x > 0, Y > 0 and 2x + y = 6, then the maximum value of XY is?
A.3 B.2/9 C.4 D.5

2x+y=6>/=2(2xy)1\2
3>\=(2xy)1\2
9>\=2xy
9\2>\=XY
(xy)max=9\2

If 2x + y = 6 and x > 0, Y > 0, then the maximum value of XY is?

2x+y=6≥2√(2xy)
2√(2xy)≤6
√(2xy)≤3
√(xy)≤3√2/2
Because x > 0, Y > 0
So XY ≤ 9 / 2

If 2x + y = 6. X > 0, Y > 0, find the maximum value of XY

4.5
2X + Y > = 2XY under double root sign
6> = 2XY under double root sign
Simultaneous square
36>=8xy
First floor is (6 - y) * y / 2 as a quadratic function, and then find the maximum, the method is good

It is known that the line y = - x + 1 intersects the ellipse at two points a and B
(1) If the eccentricity of the ellipse is √ 3 / 3 and the focal length is 2, find the length of line ab
(2) If the vector OA and the vector ob are perpendicular to each other (o is the origin of the coordinate), when the eccentricity a of the ellipse belongs to [1 / 2, (radical 2) / 2], the maximum length of the major axis of the ellipse is obtained
I'm a liberal arts student. I'll explain it in detail,

Without conditions, the focus should be on the x-axis
(1) Centrifugation e = C / a = √ 3 / 3 = 1 / √ 3
∵ c=1,∴ a=√3
∴ b=√2
The equation is X & # 178 / 3 + Y & # 178 / 2 = 1
(2) Let a (x1, Y1), B (X2, Y2)
Substituting y = - x + 1 into B & # 178; X & # 178; + A & # 178; Y & # 178; = A & # 178; B & # 178;
∴b²x²+a²(1-x)²=a²b²
∴(a²+b²)x²-2a²x+a²(1-b²)=0
Using Veda's theorem
∴x1+x2=2a²/(a²+b²),x1*x2=a²(1-b²)/(a²+b²)
∴ y1y2=(-x1+1)*(-x2+1)=x1x2-(x1+x2)+1=[a²(1-b²)-2a²+a²+b²]/(a²+b²)
∴ y1y2=b²(1-a²)/(a²+b²)
OA and ob are perpendicular to each other
∴ x1x2+y1y2=0
∴ a²(1-b²)+b²(1-a²)=0
That is, a & # 178; + B & # 178; = 2A & # 178; B & # 178;
∴ a²+a²-c²=2a²(a²-c²)
∴ 2a²=(2a²-c²)/(a²-c²)
Divide the fraction by a and 178 at the same time;
∴ 2a²=(2-e²)/(1-e²)=1+1/(1-e²)
∵ e∈[1/2,(√2)/2]
∴ e²∈[1/4,1/2]
∴ 1-e²∈[1/2,3/4]
∴ 1/(1-e²)∈[4/3,2]
∴ 1+1/(1-e²)∈[7/3,3]
The maximum value of ﹥ 2A & ﹥ 178; is 3
The maximum value of a is √ (3 / 2) = √ 6 / 2
The maximum value of long axis length is √ 6

It is known that P is a point in the third quadrant of the ellipse x245 + y220 = 1, and it is perpendicular to the two focal lines. If the distance between P and the line 4x-3y-2m + 1 = 0 is not greater than 3, then the value range of real number m is ()
A. [-7，8]B. [−92，212]C. [-2，2]D. （-∞，-7]∪[8，+∞）

∵ the two focus coordinates of the ellipse x245 + y220 = 1 are (- 5,0) (5,0), and P (x, y) (x ＜ 0, y ＜ 0) and the two focus lines are perpendicular to each other, ∵ YX + 5 · & nbsp; YX − 5 = − 1, that is, y2 = 25-x2

Given the point a (1,1), and F1 is the left focus of the ellipse X29 + Y25 = 1, P is any point on the ellipse, then the minimum value of | Pf1 | + | PA | is ()
A. 6-2B. 6+2C. 10D. 2

So, |pf1 124124124124124124124124124124\124124124124124124\124124124124\124\124\\\\124\pf2 = 2A = 6 = 6, then, \\\\\\\\\\\\\\\\\\-2

Let f (x) = lnx-2ax. (1) if the tangent of the image of function y = f (x) at point (1, f (1)) is a straight line L, and the line L is tangent to the circle (x + 1) 2 + y2 = 1, find the value of a; (2) when a > 0, find the monotone interval of function f (x)

(1) Therefore, the slope of the line passing through (1, f (1)) is 1-2a, and f (1) = - 2A. Therefore, the equation of the line passing through (1, f (1)) is y + 2A = (1-2a) (x-1)