If the distance from a point m on the parabola x ^ 2 = - 16y to the focus is 6, then the coordinate of M is

If the distance from a point m on the parabola x ^ 2 = - 16y to the focus is 6, then the coordinate of M is

2p=16,p=8,p/2=4.
Focus f (0, - P / 2), namely f (0, - 4)
Let m (x, y)
|FM|=√[(x-0)^2+(y+4)^2]=6.
Square of both sides: x ^ 2 + y ^ 2 + 8y + 16 = 36
-16y+y^2+8y-20=0.
y^2-8y-20=0.
(y-10)(y+2)=0.
Y-10 = 0, y = 10 (this point is not on the parabola, round off
∴y=-2.
Substituting y = - 2 into x ^ 2 = - 16y, x ^ 2 = 32, x = ± 4 √ 2
The coordinates of m point are M1 (- 4 √ 2, - 2), M2 (4 √ 2, - 2)

The distance from a point m to the focus on the parabola x ^ 2 = 4Y is equal to 3. Find the distance from the point m to the parabola collimator, and find the ordinate and abscissa of the point M

According to the second definition of parabola (the ratio of fixed point to fixed line is 1), the distance from a point m on the parabola x ^ 2 = 4Y to the focus is equal to 3,2
It can be seen that the fixed straight line is y = - P / 2 = - 2 / 2 = - 1, so the ordinate is 2, and y = 2 is substituted into the parabola to get x = - 2 times root 3 or 2 times root 3

The distance from a point m on the parabola x ^ 2 = 4Y to the focus is 8. What is the distance from point m to the collimator? What is the ordinate of point m

The focal coordinate (0,1) and the Quasilinear equation are y = - 1
Let m coordinate be (a, a ^ 2 / 4)
The distance from m to the focus is 8, with the relation a ^ 2 + (a ^ 2 / 4-1) ^ 2 = 64
The simplified equation is: (a ^ 2-36) (a ^ 2 + 28) = 0
The solution is a = 6 or - 6
The ordinate of point m is: 9
The distance to the guide line is 9 - (- 1) = 10