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A math problem in high school, thank you!
Solution:

Obviously, we can know that the slopes of AB and CD both exist, otherwise one of them has only one intersection with parabola.

And focus coordinate (a, 0) and directrix x =-a.

So let AB:y=k(x-a)

Because AB and CD are vertical, the slope is negative reciprocal.

Then CD: y =- 1/k * (x-a)

Combined with parabolic equation:

About AB: k 2 * x 2-(2 * a * k 2+4a) * x+a 2 * k 2 = 0.

On the CD: x 2-(4 * a * k 2+2a) * x+a 2 = 0.

Vieta's theorems are: xa+XB = (2 * a * k 2+4a)/k 2, xc+xd = 4 * a * k 2+2a.

Since AF, BF, CF and DF are over-focused, they are defined by parabolas as follows:

AF=xA+a,

BF=xB+a,

CF=xC+a,

DF=xD+a

therefore

|ab|+|cd|=xa+a+xb+a+xc+a+xd+a=8a+4ak^2+4a/k^2

According to the basic inequality: the minimum value is 16a (if and only if k=2√a, take the equal sign).