If the square of (3x-4y-10) + |5x + 6y-4 | = 0, then x =, y=

If the square of (3x-4y-10) + |5x + 6y-4 | = 0, then x =, y=

The title means (3x-4y-10) = 0;
|5x+6y-4|=0；
x=2,y=-1

If point P is any point on the curve y = 2 - ㏑ 2x, then the minimum distance from point P to the straight line y = - x is

Let P (x, 2-ln2x) be the distance from it to the line x + y = 0
d=|x+2-ln2x|/√2,
Let f (x) = x + 2-ln2x, x > 0, then
f'(x)=1-2/(2x)=1-1/x,↑
00,f(x)↑,
∴f(x)|min=f(1)=3-ln2>0,
∴d|min=(3-ln2)/√2.

Find a point P on the curve C: 2x-y & # 178; - 1 = 0, so that the distance from P to the straight line L: 2x-y + 3 = 0 is the shortest, and find the shortest distance

x=(y²+1)/2
P[(a²+1)/2,a]
Then d = | A & # 178; + 1-A + 3 | / √ (2 & # 178; + 1 & # 178;)
=[(a-1/2)²+15/4]/√5
So when a = 1 / 2, D is the minimum
So p (5 / 8,1 / 2), D min = (15 / 4) / 5 = 3 / 4

What is the shortest distance from the point on the curve y = x ^ 4 to the line x-2x-1 = 0
Change to x-2y-1 = 0

Make a tangent parallel to x-2y-1 = 0, slope = 1 / 2
Then the distance between two straight lines is the minimum distance
So y '= 4x ^ 3 = 1 / 2
x=1/2
y=x^4=1/16
Tangent point (1 / 2,1 / 16)
That is to find the distance from x-2y-1 = 0
So = root 5 / 8

What is the shortest distance from a point on the curve y = ln (2x-1) to the line 2x-y + 3 = 0______ ．

Because the slope of the straight line 2x-y + 3 = 0 is 2, let y '= 22x − 1 = 2, the solution is: x = 1, and substitute x = 1 into the curve equation to get: y = 0, that is, the tangent slope of the curve passing (1,0) is 2, then the distance from (1,0) to the straight line 2x-y + 3 = 0 d = | 2 + 3 | 22 + (− 1) 2 = 5, that is, the shortest distance from the point on the curve y = ln (2x-1) to the straight line 2x-y + 3 = 0 is 5

What is the shortest distance from a point on the curve y = ln (2x-1) to the line 2x-y + 3 = 0______ ．

Because the slope of the line 2x-y + 3 = 0 is 2, let y ′ = 22x − 1 = 2, the solution is: x = 1, and substitute x = 1 into the curve equation to get: y = 0, that is, the tangent slope of the curve passing (1,0) is 2, then the distance from (1,0) to the line 2x-y + 3 = 0 d = | 2 + 3 | 22 + (− 1) 2 = 5, that is, the point on the curve y = ln (2x-1) to the line 2x -

Find the point on the curve y = 4-x ^ 2 which is closest to the fixed point P (0,2)
As above

Take any point on the curve as (x1, Y1), then the square of the distance is d ^ 2 = X1 ^ 2 + (y1-2) ^ 2 = X1 ^ 2 + (2-x1 ^ 2) ^ 2
Let x ^ 2 be a variable, that is, X1 ^ 4 + 4-3x1 ^ 2, X1 ^ 2 = t (T > = 0). Remember the answer, (√ 6 / 2, √ 10 / 2)
I hope it can be used. The answer is not calculated carefully

Let p be a moving point on the curve y2 = 4 (x-1), then the minimum value of the sum of the distance from point P to point (0,1) and the distance from point P to Y-axis is ⊙___ ．

The image of y2 = 4 (x-1) is a parabola with Y-axis as the guide line and (2,0) as the focus. When the point P is the intersection of the connecting line between (0,1) point and (2,0) point and the parabola, the distance sum is the minimum, and the minimum value is: (2-0) 2 + (0-1) 2 = 5

If the point P is any point on the curve y = x ^ 2-inx, then the minimum distance between the point P and the line y = X-2
Why? How to see if there is an intersection between a line and a curve
For example, another question [what is the shortest distance from the point on the curve y = ln (2x-1) to the line 2x-y + 3 = 0]

The shortest distance from point P to the straight line can be seen as translating the straight line to be tangent to the curve, changing the intercept of the new line, keeping the slope unchanged, or 1. The derivative of the curve is the slope of the tangent, so y "= 2x-1 / x = 1, so x = 1 or x = - 1 / 2 (rounding), so p (1,1), so the formula of the distance from the root point to the straight line is d = root 2
Whether there is an intersection between a straight line and a curve is mainly to substitute the linear equation into the curve, and see how many solutions of the equation are. If B ^ 2-4ac = O has an intersection, if it is greater than 0, there are two intersections
The following problem is the same solution as the first one!

If the distance between the point P on the curve y = 16 / X and the line 4x + y + 9 = 0 is the shortest, the coordinates of the point P are obtained

Let 4x + y + Z = 0. Then this straight line is parallel to the problem. Substituting y = - 4x-z into the curve, we get - 4x square - ZX-16 = 0, Δ = 0, we get z = 16. Then 4x + y + 16 = 0, together with the curve equation, we get x = 2, y = 8