Find a point m on the parabola x ^ 2 = 1 / 4Y, so that the distance from m to the straight line y = 4x-5 is the shortest

The M coordinate is (a, 4A ^ 2) 4x-y-5 = 0m to the distance of the straight line = 124;4a-4a-4a ^ 2-2-2-5 | / is to find the minimum value of the minimum value of the minimum value of the minimum value of the minimum value of the minimum value of the minimum value of the minimum value of the minimum value of the small-4a-4a-4a ^ 2-4-4a ^ 2-2-2-4 (A-1 / 2) ^ 2 (A-1 / 2 (A-1 / 2) ^ 2 (A-1 / 2) / 4 (A-1 / 2) (4 (A-1 / 2) ^ 2 + 2 + 4 124; so the | 4A ^ 4A ^ 4A ^ 4A ^ 4A ^ 4A ^ 4A + 5 \124; ^ 44444; ^ 4; ^ 4; ^ 4; ^ 4/ 2,1)

The distance from a point P on the parabola y ^ 2 = 4x to the directrix is D1, and the distance from a point P to the straight line x + 2y-12 = 0 is D2. Find the minimum value of D1 + D2,

The distance from point P to the guide line is equal to the distance from point P to the focus, which is PF
The distance from point P to line x + 2y-12 = 0 is set as PQ
Then D1 + D2 = PQ + PF ≥ QF, when it is equal to QF, there is a minimum value
QF is the distance from the focus (1,0) to the straight line, which is 9 / 5 * root 5

Given that point P is any point on parabola y = (1 / 4) x (2) + 1, note that the distance between point P and X axis is D1, and the distance between point P and point F (0,2) is D2
1) Conjecture and prove the size relation of D1 and D2.
(2) If the straight line pf intersects the parabola at another point Q (different from point P), try to judge the position relationship between the circle with the diameter of PQ and the X axis, and explain the reason.
I can write the first question

Y = x ^ 2 / 4 + 1 x ^ 2 = 4 (Y-1) vertex (0,1) focus f (0,2) collimator y = 0, that is, x-axis is defined by the second definition of parabola D1 = D22, tangent to X-axis pf = P, distance from X-axis QF = q, distance from X-axis PQ = distance from P to X-axis + distance from Q to X-axis PQ / 2 = (distance from P to X-axis + distance from Q to X-axis) / 2, that is, in PQ

Given that point P is a point on parabola y2 = 4x, let the distance from point P to the Quasilinear of parabola be D1, and the distance from point P to moving point Q on circle (x + 3) 2 + (Y-3) 2 = 1 be D2, then the minimum value of D1 + D2 is ()
A. 3B. 4C. 5D. 33+1

Connect the focus and center of the parabola. From the definition of the parabola, we know that the length of the line between the two points minus the radius of the circle, which is the minimum distance I asked, ∵ the focus of the parabola is (1,0) the center of the circle is (- 3,3) ∵ the minimum value of D1 + D2 is (− 3 − 1) 2 + (0 − 3) 2 − 1 = 4, so choose B

Given that point P is a point on parabola y2 = 4x, let the distance from point P to the Quasilinear of parabola be D1, and the distance from point P to moving point Q on circle (x + 3) 2 + (Y-3) 2 = 1 be D2, then the minimum value of D1 + D2 is ()
A. 3B. 4C. 5D. 33+1

The vertex of parabola C is at the origin of coordinate, the symmetry axis is y-axis, and the shortest distance between the upper moving point P of parabola C and the straight line L: 3x + 4y-12 = 0 is 1

The slope k of the straight line L: 3x + 4y-12 = 0 = - 34, the intercept on the Y axis: 3, if the opening of the parabola is upward, it will intersect with the straight line L, and the shortest distance will not be equal to 1, so the parabola opening is downward, let its equation be: x2 = - 2PY, (P > 0) the point of the shortest distance from the parabola to the straight line L is the tangent point of the tangent m of the parabola parallel to L, and the shortest distance is the distance from the tangent to L The equation is 3x + 4Y + q = 0, let the distance between M and l be | Q + 12 | 9 + 16 = 1, then q = - 7 or - 17 can be obtained, and q = - 17 is below L, rounding off. So m: 3x + 4y-7 = 0. Combined with the parabolic equation x2 = - 2PY, we can get 2x2-3px-7p = 0, only one common point, △ = 9p2 + 56p = P (9p + 56) = 0, we can get P = 569, so the equation of C is: x2 = 2 (- 569) y, that is 9x2 + 112Y = 0

Find a point on the parabola y = 4x square, so that the distance from this point to the straight line y = 4x-5 is the shortest

Let the point be a (a, b), B = 4A ^ 2
A(a,4a^2)
y = 4x -5,4x -y -5 = 0
Distance from a to straight line d = | 4A - 4A ^ 2 - 5 | / √ [4 ^ 2 + (- 1) ^ 2] = | 4A ^ 2 - 4A + 5 | / √ 17 = | (2a-1) ^ 2 + 4 | / √ 17
When a = 1 / 2, the distance from a to the straight line is the shortest, a (1 / 2,1)

Find a point m on the parabola x2 = 1 / 4Y, so that the distance from m to y = 4x-5 is the shortest

If the distance between the point m on the parabola x ^ 2 = (1 / 4) y and y = 4x-5 is the shortest, then the tangent passing through the point m on the parabola x ^ 2 = (1 / 4) y is parallel to the straight line y = 4x-5
y=4x+b
Substituting x ^ 2 = (1 / 4) y, we get
x^2=(1/4)y=(1/4)*(4x+b)
x^2-x-(5/4)b=0
If the line passing through the point m y = 4x + B is tangent to the parabola x ^ 2 = (1 / 4) y, then the discriminant of the above equation △ = 0, i.e
△=(-1)^2-4*1*[-(5/4)b]=0
b=-1/5
The line passing through point m y = 4x-1 / 5
Now the problem is to find the distance between two straight lines y = 4x-5 and y = 4x-1 / 5
y=4x-5
x=0,y=-5,N(0,-5)
The distance from point n (0, - 5) to y = 4x-1 / 5
L=|4*0-(-5)-1/5|/√17 =4.8/√17
The shortest distance from m to y = 4x-5 = 4.8 / √ 17

The distance from point P (- 1,2) to line 3x-4y + 1 = 0 is?

The distance from point P (- 1,2) to line 3x-4y + 1 = 0 is d = | - 3-8 + 1 | / √ (3 ^ 2 + 4 ^ 2) = 10 / 5 = 2
If you don't understand, please ask, I wish you a happy study!

On the straight line 3x-4y-27 = 0, the coordinates of the nearest point to point P (2,1) are ()
A. （5，-3）B. （9，0）C. （-3，5）D. （-5，3）

Because the slope of the known line 3x-4y-27 = 0 is 34, so the slope of the line passing through P is − 43, and P (2,1), then the equation of the line is: Y-1 = − 43 (X-2), that is 4x + 3y-11 = 0, and

Find a point m on the parabola x ^ 2 = 1 / 4Y, so that the distance from m to the straight line y = 4x-5 is the shortest