Let's find the coordinates of d first. The abscissa is obviously -6, and the ordinate is F (-6) = A (-6+6) 2-3 =-3. So we get D(-6, -3).
If E divides DN equally, then EN=2DE=6, and then we get n (-6,6).
Make a circle C: (x+6) 2+(y-3/2) 2 = 8 1/4 with the midpoint of DN (assuming point F) as the center and1* dn as the radius. If it intersects with the parabola in the second quadrant, there is such a thing.
Now find the value of a (in fact, I don't quite understand the meaning of AE2=3ED here, so I'll give you two understandings. ) Because 2AE=3DE=9, AE=9/2. In this way, a (-3/2,0) can be easily obtained, and then it can be brought into the analytical formula.
0=a(-3/2+6)^2-3
a*4.5^2=3
a = 3/20.25 = 12/8 1 = 4/27
Next, the equations of circle and parabola are combined, if X is in X.
(x+6)^2+(y-3/2)^2=8 1/4 - a
y=4/27*(x+6)^2-3 - b
Simplified formula a: y = [81/4-(x+6) 2]1/2. +3/2-C。
Y =-[81/4-(x+6) 2]1/2.+3/2 dimension
B and c are equal, let (x+6) 2 be m (m >; 0), there are:
4/27m-3=(8 1/4-m)^ 1/2+3/2
(4/27m) 2- 1/3m = 0 (square at the same time, then move to the same side)
M=0 or 9/4, when x=-6,-15/2, 9/2, (one of them is point D, and the other two are in the third quadrant, so they are not in the range, so they don't exist in this case).
Then put b and d together:
4/27m-3=-(8 1/4-m)^ 1/2+3/2
(4/27m) 2-7/3m+8 1/2 = 0 (obviously, (7/3) 2-4 * 4/27 * 8 1/2
Therefore, there is no such m-point.
Method 2 (Analysis):
Let FM= 1/2DN=9/2, whereas AE=9/2 (F is the midpoint of DN).
It can be easily understood from the trilateral relationship of the triangle that when m is in the second quadrant, FM >；; AM & gt9/2, so there can't be such an m-point.
Method 3: The understanding is different from that of Method 1.
In the past, when 2AE=3DN, it is now understood as AE 2 = 3dn. AE = 3 (this is actually much simpler! ! ! )
The previous steps are inconvenient, starting with finding the value. In this case, a=3 is obtained by the same method, and the analytical formula is:
Y = 3 (x+6) 2-3,-one
The equation of the circle remains unchanged: (x+6) 2+(y-3/2) 2 = 81/4-b.
Simplify b to get y = 3/2+-(81/4-m) (1/2)-.
Meanwhile, a and c: 3m-3 = 3/2+-(81/4-m) (1/2) (x
m- 1= 1/2+-(9/4-m)^( 1/2)
(m-3/2)^2=9/4-m
m^2-3m+m=0
m = 0,2
When m=0, the method 1 has been proved to be invalid.
When m=2, that is, X = 6+-2 (1/2), (not in the range of X,)
So the evidence does not exist.
Okay, that's enough. I hope it helps you. If you have any questions, just ask me and I'll try my best to answer them.