Using Pythagorean Theorem: Get
x^2+(y-5)^2+(x+2)^2+(y-6)^2=(2-0)^2+(6-5)^2
x^2+y^2- 1 1y+2x+30=0
(x+ 1)^2+(y-( 1 1/2))^2=5/4
This is the root circle with the center of (-1, 1 1/2) and the radius of half.
2. First: straight line L: (2m+1) x+(m+1) y = 7m+4 (m belongs to r).
Is a straight line through a fixed point.
Find this fixed point:
(2m+ 1)x+(m+ 1)y=7m+4
m(2x+y)+x+y=7m+4
As long as 2x+y=7
x+y=4
To get (3, 1), the straight line l: (2m+1) x+(m+1) y = 7m+4 (m belongs to r) is a straight line passing through (3,1).
Obviously this is in the circle;
Now the problem becomes the shortest chord length of a straight line L passing through a point on a circle cut by a circle C.
Obviously, when the straight line connecting this point and the center of the circle is perpendicular to the straight line, the chord is the shortest.
Because if the connecting line between the point and the center of the circle is not perpendicular to the straight line
Then the straight line perpendicular to the center of the straight line and the connecting line between the point and the center of the straight line are the relationship between the right angle and the hypotenuse.
The connecting line between the point and the center of the circle is the largest.
Obviously, the longer the line perpendicular to the center of the line, the shorter the chord, because the hypotenuse of the right triangle they form is the radius r, which is certain.
So: Find the slope of the connecting line between this point and the center of the circle:
(2- 1)/( 1-3)=- 1/2
So the slope of L: (2m+1) x+(m+1) y = 7m+4 (m belongs to R) =-1-1/2 = 2 (vertical relationship).
So:
-(2m+ 1)/(m+ 1)= 2m =-3/4
The shortest length of the chord = √( R2- the square of the connecting line between the point and the center of the circle)
Square of the distance from the point to the center of the circle: (1-3) 2+(2- 1) 2 = 5.
Minimum length = √ (5 2-5) = √ 20 = 4 √ 5
3. Take the symmetry point A'(0, 1) of A about the line 2x-y-4=0.
|PA-PB|=|PA'-PB|≤A'B=3√2 (the difference between two sides of a triangle is smaller than the third side).
When p is located on the extension line of A'B, take the equal sign.
A'B equation: y=x+ 1, and 2x-y-4=0, p (5,6).
-& gt； When the p coordinate is (5,6), the maximum distance difference =3√2.
4. Let A(x 1, y 1) B(x2, y2).
The equation of AB is: y=[ (root number 3)/3]x+p/2.
The focus is (0, p/2). The alignment equation is y=-p/2.
W is perpendicular to the directrix of A and B, and the crosshairs are at C and D. 。
According to the definition: |AF|=|AC|=y 1+p/2=[ (root number 3)/3]x 1+p,
|BF|=|BD|=y2+p/2=[ (root number 3)/3]x2+p,.
|AF|/|BF|={[ (root number 3)/3]x 1+p}/{[ (root number 3)/3]x2+p} (1).
Then find the abscissa of the intersection and solve the equation: x 2 = 2p {[ (root number 3)/3]x+p/2}.
Namely: x 2-2p [(root number 3)/3] x-p 2 = 0.
Solution: x 1=[ (radical number 3)/3- (radical number 2)/3] p =-[ (radical number 3)/3)/3] p.
X2=[ (root number 3)/3+ (root number 2 3)/3] p = (root number 3) p.
Substitute (1) to get:
| AF |/| BF | =(- 1/3+ 1)/( 1+ 1)= 1/3
5.a^2=4,b^2=3
Then c 2 = 1
e=c/a= 1/2
Distance from MF/M to the right line = 1/2.
Distance from m to right directrix =2MF
Right directrix x = a 2/c = 4
Distance from p to right alignment line =4- 1=3.
Align PQ vertically to the right,
Then when m is the intersection of PQ and ellipse, the sum of distances is the smallest.
So M(x,-1)
Substitution, m (2 √ 6/3,-1)