If the sum of the distance from a point P to a (3, - 1) on the parabola y ^ 2 = x and the distance to the focus f is the smallest, what is the coordinate of point P

If the sum of the distance from a point P to a (3, - 1) on the parabola y ^ 2 = x and the distance to the focus f is the smallest, what is the coordinate of point P

The distance between P and focus f = the distance between P and guide line x = - 1 / 4
So I'm going to do ab. the intersection of the vertical collimator and the parabola is p
Then AB is y = - 1
So x = 1
P(1,-1)

The shortest distance from the point on the parabola y = X2 to the line x-y-2 = 0 is ()
A. 2B. 728C. 22D. 1

Let d = | m − M2 − 2 | 2 = | (m − 12) 2 + 74 | 2 be the distance from any point m (m, M2) m on the parabola to the straight line x-y-2 = 0. According to the properties of quadratic function, when m = 12, the minimum distance D = 728

The shortest distance from a point on the parabola y = x ^ 2 to the straight line X-Y = 2?

Let the point be (a, b) on y = x & sup2;, B = A & sup2; (a, a & sup2;) to x-y-2 = 0, distance = | A-A & sup2; - 2 | / √ (1 & sup2; + 1 & sup2;) = | A & sup2; - A + 2 | / √ 2 = | (A-1 / 2) & sup2; + 7 / 4 | / √ 2, molecular minimum = 7 / 4, so the shortest distance = (7 / 4) / √ 2 = 7 √ 2 / 8

Find the shortest distance between parabola y = x ^ 2 and straight line x-y-2 = 0

Solution: translate the line x-y-2 = 0 until it is tangent to the parabola y = x ^ 2. At this time, the distance from the tangent point to the line x-y-2 = 0 or the distance between two parallel lines is the shortest distance from the parabola y = x ^ 2 to the line x-y-2 = 0. Let the line tangent to the parabola y = x ^ 2 be X-Y + M = 0