Several areas of inverse proportion function

Several areas of inverse proportion function

1. When P point is fixed, find the linear equation with the smallest area
Given p (m, n), let the linear equation be y = KX + B, and the coordinates of the intersection of x-axis and y-axis with a and B are (- B / K, 0), (0, b)
Substituting point P, n = MK + B, k = (N-B) / m
Connecting OP, the area of triangle AOB = the area of triangle AOP + the area of triangle BOP
The heights of the two triangles are n, m, bottom OA = - B / K, OB = B
So the area of triangle AOB is:
S=1/2*(-b/k)*n+1/2*b*m
Substitute K into (because KB is two unknowns, remove one)
S=-1/2*bmn/(n-b)+bm/2
=bm/2*[1-n/(n-b)]
=bm/2*b/(n-b)
=m/2*b^2/(n-b)
Let N-B = x, then ob = b = x + n
S=m/2*(X+n)^2/X
=m/2*(X^2+2Xn+n^2)/X
=m/2*(X+n^2/X+2n)
Find the minimum value of S, that is, find the minimum value of (x + n ^ 2 / x)
According to the formula a + b > = 2 times the root sign (AB), (if and only if a = B, the equal sign holds)
X + n ^ 2 / x > = 2n, when x = n ^ 2 / x, that is, x = n, x + n ^ 2 / x = 2n
So smin = m / 2 * 4N = 2Mn
Then ob = b = x + n = 2n,
k=(n-b)/m=-n/m,OA=2m
So the linear equation is y = - NX / M + 2n, and the point P is in the middle of the line segment
2. If the line is known, the method is similar to that above
Just let m, n be represented by K, P
If B = 2n, k = - N / m, then n = B / 2, M = - B / 2K
That's the midpoint