The distance from the focus of the parabola y = 4x2 to the collimator is______ ．

The distance from the focus of the parabola y = 4x2 to the collimator is______ ．

Parabola y = 4x2, that is, X2 = 14y, P = 18, that is, the distance from the focus to the collimator is equal to 18, so the answer is 18

(2012. Ziyang three mode) the distance from the focus of parabola y2=4x to the alignment is ().
A. 1B. 2C. 4D. 8

According to the meaning of the title, we can know that the focal point F (1,0), the equation x = - 1, and the distance between the focal point and the directrix is 1 + 1 = 2

The distance from the focus of the parabola y = 4x2 to the straight line y = x is ()
A. 22B. 2C. 232D. 216

Because the parabola y = 4x2 can be transformed into: x2 = 14y. 2p = 14 ﹥ P = 18 ﹥ P2 = 116. So we can get its focus coordinate as: (0116). So the distance between the point (0116) and the line X-Y = 0 d = | 0 − 116 | 12 + (− 1) 2 = 1162 = 232

Given the center m (1,1) of the square ABCD, the equation of the side where the CD is located is x + 3y-5 = 0. Find the linear equation of the other three sides

The CD equation is x + 3y-5 = 0 & # 8658; the distance from m to CD = | 1 + 3-5 | / √ ((1 ^ 2) + (3 ^ 2)) = √ (10) / 10
The analytical formula is y = - 1 / 3x + 5 / 3,
AB∥DC⇒k1=k2
Let AB analytic formula y = - 1 / 3x + B & # 8658; X + 3y-3b = 0 & # 8658; | 1 + 3-3b | / √ ((1 ^ 2) + (3 ^ 2)) = √ (10) / 10
4-3b = ± 1 & ᦇ 8658; b = 1 or 5 / 3 (take B = 1)
⇒ ab equation x + 3y-3 = 0
The other two straight lines are perpendicular to CD, K1 × K2 = - 1 & # 8658; the analytic formula is y = 3x + B & # 8658; 3x-y + B = 0
|3-1 + B | √ ((3 ^ 2) + (1 ^ 2)) = √ (10) / 10 &; b = - 1 or - 3
The other two linear equations are 3x-y-1 = 0 or 3x-y-3 = 0