The shortest distance from parabola x ^ 2 = y to straight line 2x-y-4 = 0 is? The answer is (3 root 5) / 5,

The shortest distance from parabola x ^ 2 = y to straight line 2x-y-4 = 0 is? The answer is (3 root 5) / 5,

Method 1
Let the line 2x-y + C = 0 be tangent to the parabola, then the distance from the tangent point to the line 2x-y-4 = 0, that is, the distance between two parallel lines, is the shortest distance
Substituting y = 2x + C into the parabolic equation is x ^ 2-2x-c = 0,
Let the discriminant = 4 + 4C = 0, then C = - 1,
So the shortest distance is = | - 1 + 4 | / √ (4 + 1) = 3 √ 5 / 5
Method 2
Let P (x, x ^ 2) be any point of the parabola, and the distance from P to the line 2x-y-4 = 0 is
d=|2x-x^2-4|/√5=|(x-1)^2+3|/√5 ,
Because (x-1) ^ 2 + 3 > = 3, when x = 1, the shortest distance is 3 / √ 5 = 3 √ 5 / 5

Find the minimum distance between the point on the parabola y2 = 64x and the straight line 4x + 3Y + 46 = 0, and find the coordinates of this point. The derivative can not be solved

You can ask. You'll know when you draw a picture,
The minimum point is the point where the line 4x + 3Y + 46 = 0 is continuously translated to the parabola y2 = 64x, and then the first point is tangent to the parabola
The tangent of the point tangent to the parabola is the straight line 4x + 3Y + 46 = 0. Therefore, the derivative value of that point is the slope of the straight line 4x + 3Y + 46 = 0, because the slope can be known as - 3 / 4. Therefore, the derivative value of the parabola at this point is - 3 / 4. By deriving the parabola and substituting - 3 / 4 into the derivative function, the coordinates of this point can be obtained