Enumerable equation
(x0+ 1)^2=(x0- 1)^2+y0^2
= & gt4x0=y0^2
The trajectory equation is 4x = y 2.
(2)
Let the passing point equation (-1, 0) be y=k(x+ 1).
It coincides with the parabola 4x = y 2.
You can get k2 * x2+(2k 2-4) x+k2 = 0.
Vieta's theorem has.
X 1+x2=4/k^2-2
X 1*x2= 1
Then the coordinates of the midpoint (x0, y0) of the two intersections are
((x 1+x2)/2, (y 1+y2)/2) is (2/k 2-1,2/k).
A vertical line that passes through the point to form a straight line.
The vertical equation is
y-k/2=- 1/k(x-2/k^2+ 1)
If the intersection with the X axis is (2/k 2+1,0), then the point is the desired point.
There is another condition that satisfies the regular triangle, which we haven't used yet.
As we know, the height of a regular triangle is 3 to 2 times that of the root of the base.
First, calculate the side length of a regular triangle.
From Vieta's theorem, we can calculate |x 1-x2|, and the side length is k 2+1root sign multiplied by |x 1-x2|.
The solution is √ (k 2+1) √ (16/k 4-16/k 2).
Height is the distance from the intersection of the x axis to the straight line just obtained.
Yes (2/k+2k)/√ (k 2+1)
Substituting the ratio mentioned above, we can get k= plus or minus 2 by solving the equation.
Final verification
That's the idea. The calculation result may be incorrect (the number cannot be sent -v-)
(PS: LS, if the equilateral triangle condition is used, it is too troublesome to calculate the values of two intersections, and the calculation amount is too large. This kind of problem is generally considered to be done by symmetry. For example, this problem is considered from the midpoint of two intersections, and the calculation is relatively simple. Although he usually dies after going to college, high school still needs a little skill-v-)