高職數學 求救~~

(1) (y-k)^2 = 4p*(x-h)L: x = -1 => p = 1V = (1,2) => (y-2)^2 = 4*(x-1)4x = y^2 - 4y + 8
(2-1) (y-k)^2 = 4p(x-h)L: x = -1

V = (1,3) => p = 1 + 1 = 28(x - 1) = (y - 3)^28x = y^2 - 6y + 17(2-2)

V = (3,2), F = (1,2)=> p = 3-1 = 2=> L: x = 3 + p = 5=> -8(x-3) = (y-2)^2
(2-3) A=(6,4)(y-3)^2 = 4p*(x-2)(4-3)^2 = 4p(6-2)=> p=1/16(y-3)^2 = (x-2)/4
(2-4) A=(2,3), B=(-1,6), L: x=1=> (x-1)^2 = 4p(y-k)=> 1=4p(3-k), 4=4p(6-k)=> k=2, p=1/4=> (x-1)^2 = y-2

2015-02-17 08:58:44 補充：
(3) 9x^2 + 4y^2 = 36 => (x/2)^2 + (y/3)^2 = 1

A = (+-2,0)

B = (0,+-3)


(4) (x-2)^2 + y^2 = 8 => C = (2,0), r=2√2

L1: x - y + 6 = 0

Dist = |2 - 0 + 6|/√2

= 8/√2

= 4√2 > r

= 不相交

2015-02-17 13:29:36 補充：
L2: x - y + 2 = 0

Dist = |2 - 0 + 2|/√2

= 4/√2

= 2√2

= r

= 相切


L3: x - y - 1 = 0

Dist = |2 - 0 - 1|/√2

= 1/√2

= √2/2 < r

= 相交

2015-02-17 13:40:16 補充：
(5) C: x^2 + y^2 - 2x + 4y - 20 =0

P = (xo, yo) = (-2, 2)

Tangent Line:

0 = (xo + a/2)x + (yo + b/2)y + (a*xo + b*yo)/2 + c

= (-2 - 1)x + (2 + 2)y + (2*2 + 2*4)/2 - 20


= -3x + 4y - 14

2015-02-17 13:43:58 補充：
(6) L: x = 1 - 2y

C: 0 = x^2 + y^2 + 3x + y -6


= (1 - 2y)^2 + y^2 + 3(1 - 2y) + y - 6

= 5y^2 - 9y - 2

= (5y + 1)(y - 2)

=> y = -1/5, 2

=> x = 1 - 2y = 7/5, -3

ans = (-3, 2), (7/5, -1/5)

2015-02-17 13:48:45 補充：
(8) (0,1),(2,2),(8,3)

y(x) = ax^2 + bx + c

y(0) = c = 1

y(2) 4a + 2b + 1 =2 => 4a + 2b = 1

y(8) = 64a + 8b + 1 = 3 => 32a + 4b = 1

=> a = -1/24, b = 7/12

=> 24y = -x^2 + 14x + 24

2015-02-17 13:49:40 補充：
(9) y^2 = 4px = 6x => p = 3/2

L: x = -p = -3/2

F = (p, 0) = (3/2, 0)

Lf = 2*√(4px)

= 2*√(4p^2)

= 2*2p

= 4p

= 4*3/2

= 6

建議把標題改成國二數學~

大師，我看過課本了
題目不一樣
小弟數學基礎不好
只能跑來這求救

都是課本上會有的基本題,翻課本應該會比較快.