Given that the distance from the point P on the parabola x2 = 4Y to the focus is 10, then the coordinate of point P is 0______ ．

Given that the distance from the point P on the parabola x2 = 4Y to the focus is 10, then the coordinate of point P is 0______ ．

According to the definition of parabola, the distance from the point P to the focus is equal to the distance from the point P to the collimator, YP + 1 = 10, YP = 9, x = ± 6, P, and the point coordinate is (± 6, 9), so the answer is: (± 6, 9)

The distance from a point P to the focus on the parabola x ^ 2 = 8y is 10. Find the distance from P to the directrix and the coordinates of point P

Answers;
x²=8y
Focus f (0,2), collimator y = - 2
Using the definition of parabola, the distance from P to the focus is equal to the distance from P to the directrix
The distance from P to the guide line is 10
The ordinate of P is 10 + (- 2) = 8
Substituting into the parabolic equation x & # 178; = 8y
Then x & # 178; = 64, x = 8 or - 8
The coordinates of P are (8,8) or (- 8,8)

Given that the distance from the point P on the parabola x2 = 4Y to the focus is 10, then the coordinate of point P is 0______ ．

According to the definition of parabola, the distance from the point P to the focus is equal to the distance from the point P to the collimator, YP + 1 = 10, YP = 9, x = ± 6, P, and the point coordinate is (± 6, 9), so the answer is: (± 6, 9)

If the parabola x = - 4Y & # 178; the distance from the previous point m to the focus f is 1, then what is the abscissa of point m

The parabolic standard equation is
y^2=-x/4
Therefore, the Quasilinear equation is
x=1/16
The distance between M and focus f is 1
According to the definition of parabola,
The distance between M and the guide line is 1
| X-1 / 16 | = 1 and X ≤ 0
The solution is x = - 15 / 16