What is the shortest distance from the point P (1,0) to the point on the curve y ^ = 4x

What is the shortest distance from the point P (1,0) to the point on the curve y ^ = 4x

Take any point a (A & # 178 / 4, a) on the curve, so the distance from point P (1,0) to point A: D = √ [(A & # 178 / 4-1) & # 178; + (A-0) & # 178;] = (1 / 4) √ (a & # 178; + 4) & # 178; = (1 / 4) (A & # 178; + 4) because a & # 178; + 4 ≥ 4, d = (1 / 4) (A & # 178; + 4) ≥ (1 / 4) × 4 = 1, so the minimum distance is

Find a point P on the curve C: y ^ 2 = - 4x + 8 to make it the shortest distance to the straight line L: x + Y-5 = 0, and find the shortest distance

Let x + y + C = 0 be tangent to C: y ^ 2 = - 4x + 8, x + y + C = 0y ^ 2 = - 4x + 8 eliminate x to get y ^ 2-4y + (- 8-4c) = 0, discriminant = 16-4 (- 8-4c) = 0C = - 3Y = 2, so tangent equation is x + Y-3 = 0y = 2x = 1, so the shortest distance from P (1,2) is p to l: x + Y-5 = 0, distance d = | 1 + 2-5 | / radical 2 = radical 2

The minimum distance between the point on the curve y = x ^ 2 + X and the line 3x-y-3 = 0 is

I'd like to answer you, but I don't remember the formula for a long time. We know that the intersection point of the vertical line and the parabola is (x, y). Using the distance from the point to the line, we can find the equation on the parabola

How to find the asymptote of quadratic curve X ^ 2-3xy + 2Y ^ 2 + x-3y + 4 = 0?
Give the best process. Thank you

The curve can be reduced to (y + 3) ^ 2 / 24 - (x-1.5y + 0.5) ^ 2 / 6 = 1
Let m = y + 3, n = x-1.5y + 0.5
We can see that it is a hyperbolic asymptote, which can be easily solved by simplifying it

Given x > 0, Y > 0 and X + 3Y + 3xy = 15, find the minimum value of X + 3Y

3xy = 15, the solution is x + 3Y > = 6, where x = 3, y = 1, and the minimum value is 6

The curve of x ^ 2-3xy ^ 2 = 1 is

x²-3xy+y²-1=0
The discriminant △ = 9-4 > 0
So it's a hyperbola

Let K be a positive real number. If the equation kxy + xsquare-x + 4y-6 = 0 represents two straight lines, then the equations of the two straight lines are?
How come you can't see the answer as soon as you open it? Please answer again,

If x ^ 2 + X (ky-1) + 4y-6 = 0 is a straight line, then it can be decomposed into two linear factors, that is, Delta1 is a complete square formula: get (ky-1) ^ 2-4 (4y-6) = k ^ 2Y ^ 2-2y (K + 8) + 25 is a complete square formula. Delta2 = 04 (K + 8) ^ 2-100k ^ 2 = 0 is transformed into: 3K ^ 2-2k-8 = 0 (3K + 4) (K-2) = 0. Because k > 0, there can only be k = 2

Let K be a positive integer, and the equation kxy + x ^ 2-x + 4y-6 = 0 represent two straight lines. Find the equation of these two straight lines. The simpler the solution, the better

Using undetermined coefficient method
Since kxy + x ^ 2-x + 4y-6 = 0 represents two straight lines, and there is no y ^ 2 term in the equation, let
kxy+x^2-x+4y-6=0=(x+a)(x+by+c)
The results are as follows
x^2+bxy+cx+ax+aby+ac=kxy+x^2-x+4y-6
The comparison coefficient is as follows:
b=k,a+c=-1,ab=4,ac=-6
The solution is a = - 3, C = 2, or a = 2, C = - 3
So k = b = - 4 / 3 or 2
When k = - 4 / 3, the two equations are x-3 = 0, x-4 / 3Y + 2 = 0
When k = 2, the two equations are x + 2 = 0, x + 2y-3 = 0

The linear equation of x-2y + 1 = 0 with respect to x = 1 symmetry is______ ．

In the straight line x-2y + 1 = 0, take any two points (1,1), (0,12). The symmetry points of these two points about the straight line x = 1 are (1,1), (2,12) respectively. The linear equation passing through these two points is Y-1 = - 12 (x-1), that is, x + 2y-3 = 0. So the answer is: x + 2y-3 = 0

The linear equation of x-2y + 1 = 0 with respect to x = 1 symmetry is
The answer is x + 2y-3 = 0, especially the slope of that line

Analysis: find two special points on the line x-2y + 1 = 0, determine the coordinates of these two points about the symmetrical point of the line x = 1, and then determine the linear equation of the line x-2y + 1 = 0 about the symmetrical point of the line x = 1 from these two symmetrical points
Method 1
∵ the line x-2y + 1 = 0 intersects the X axis at the point (- 1,0), and the line x = 1 intersects the point (1,1)
The symmetric point of point (- 1,0) with respect to line x = 1 is point (3,0), and the symmetric point of point (1,1) with respect to line x = 1 is point (1,1)
The line x-2y + 1 = 0 passes through the point (3,0) and the point (1,1) symmetric to the line x = 1
By substituting (3,0) and point (1,1) into the undetermined analytical formula, we can get the following result:
The linear equation is: x + 2y-3 = 0
Method 2: set the point (x1, Y1) on the straight line, find its symmetric point (2-x1, Y1) about x = 1, and write (X2, Y2)
x1-2y1+1=0
Y2 = Y1, X2 = 2-x1, so X1 = 2-x2
So we have (2-x2) - 2Y2 + 1 = 0
We get - x2-2y2 + 3 = 0
x+2y-3=0