If the abscissa of a point m on the parabola y2 = 16x is 6, then the distance from m to the focus f is 0______ ．

The Quasilinear equation of parabola y2 = 16x is: x = - 4 ∵ the distance from m to focus f is equal to the distance from m to Quasilinear, the abscissa of M is 6, and the distance from m to focus f is 6 + 4 = 10, so the answer is: 10

RELATED INFORMATIONS

1. What is the distance from point (3, - 6) to the focus of the parabola y square = 12x
2. Let ABC be three different points on the parabola y2 = 4x, if the focus F of the parabola is exactly the center of gravity It is known that ABC is three different points on the parabola y ^ 2 = 4x. If the focus F of the parabola is exactly the center of gravity of the triangle ABC, then the value of AF + BF + CF is equal to?
3. Let f be the focus of the parabola y2 = 4x, and a, B, C be the three points on the parabola. If the center of gravity of the points a (1,2), △ ABC coincides with the focus F of the parabola, Then the equation of the line where the BC edge is located is
4. Hyperbola 25 / x square - 9 / y square = 1 any point P is closer to this hyperbola, the distance from one focus is 2, find the distance from point P to another focus
5. Let p be a point on the hyperbola (x ^ 2 of 16) - (y ^ 2 of 9) = 1, and the distance from P to one focus of the hyperbola is 10, then what is the distance from P to another focus
6. On hyperbola x ^ 2 / 16-y ^ 2 / 9 = 1, if the distance from a point P to a focus is 12, then the distance from point P to another focus is?
7. The distance from the point P on the hyperbola x ^ 2 / 16-y ^ 2 / 9 = 1 to the right collimator is 12.5, and the distance from P to the right focus is calculated
8. If the distance from a point P to one focus on the hyperbola X225 − Y29 = 1 is 12, then the distance from it to the other focus is 12______ ．
9. The distance between the intersection point of the vertical line and the hyperbola and the two focal points can be obtained by making the x-axis vertical line between the hyperbola (xsquare) / 144 - (ysquare) / 25 = 1 and the focal point Through a focus of hyperbola (xsquare) / 144 - (ysquare) / 25 = 1, make a vertical line of X axis, and find the distance from the intersection of the vertical line and hyperbola to the two focuses A = 2x root 5, through the point a (- 5,2), the focus is on the X axis, find the standard equation of hyperbola
10. Make a vertical line of X axis through a focus of hyperbola x square / 144-y square / 25 = 1, and calculate the distance from the intersection of the vertical line and hyperbola to the two focuses
11. The square of parabola y = 2px (P > 0), the distance from a point m to the focus is a (a > P / 2), then what is the distance from point m to the collimator? The abscissa of point m is If the distance between a point m and the focus on the parabola y = 2px (P > 0) is a (a > P / 2), what is the distance between the point m and the collimator? What is the abscissa of the point m?
12. If the distance from a point m on the parabola y = 4x2 to the focus is 1, then the ordinate of point m is () A. 1716B. 1516C. 78D. 0
13. If the distance from a point m on the parabola y = 4x2 to the focus is 1, then the ordinate of point m is () A. 1716B. 1516C. 78D. 0
14. If the distance from a point m on the parabola y = 4x2 to the focus is 1, then the ordinate of point m is () A. 1716B. 1516C. 78D. 0
15. If the distance from a point m on the parabola y = 4x2 to the focus is 1, then the ordinate of point m is () A. 1716B. 1516C. 78D. 0
16. The distance from one focus to two focuses on the ellipse is D1 D2, and the focal length is 2C. If D1 2C D2 is an arithmetic sequence, the eccentricity of the ellipse can be calculated
17. It is known that P is a point on the hyperbola x ^ 2 / 4-y ^ 2 / b ^ 2, F1 and F2 are the left and right focal points, the three sides of ⊿ P F1F2 grow into an arithmetic sequence, and ∠ F1 P F2 = 120 to calculate the value of E E is the eccentricity-
18. The two foci F1, F2, P of hyperbola (x ^ 2) / 4 - (y ^ 2) / (b ^ 2) = 1 (B ∈ n *) are a point on the hyperbola, / op / < 5, / Pf1 /, / F1F2 /, / PF2 / are in equal proportion sequence, and the equation of the hyperbola is obtained
19. The two focuses of hyperbola x ^ 2 / 4-y ^ 2 / b ^ 2 = 1 are F1 and F2, and point P is on the hyperbola, if | Pf1 | F1F2 | PF2 | is an arithmetic sequence And | 0P | = 5, then B ^ 2=
20. The two foci F1, F2 and P of hyperbola x2 / 4-y2 / B2 = 1 are the points on the hyperbola, Pf1, F1F2 and PF2 form an arithmetic sequence, and op = 5, then B ^ 2 =?