Re: [多變] 求切面的點
※ 引述《stitchcca (阿迪)》之銘言：
: Find the points on the hyperboloid x^2 + 4y^2 - z^2 = 4
: where the tangent plane is parallel to the plane 2x + 2y + z = 5.
: 答案是 (2,1/2,-1),(-2,-1/2,1)
: 請問這題要怎麼求出答案
: 謝謝~
Set f(x,y,z) = x^2 + 4y^2 -z^2 - 4 =0 ,
g(x,y,z) = 2x + 2y + z =5
▽f = ( 2x , 8y , -2z )
▽g = ( 2 , 2 , 1 )
if 兩個平面平行 兩個平面各自的法向量也會平行
(2x , 8y , -2z ) =a ( 2 , 2 , 1)
where a 長度因子
2x = 2a , 8y = 2a , -2z = a
x = a , y = a/4 , z =-a/2
put they in to f , slove a=?
2 2 2
a + a /4 - a / 4 = 4 a= 2 or -2
so (x,y,z) = (2 , 1/2 , -1 ) or (-2 , -1/2 , 1)
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06/26 10:27, , 1F
06/26 10:27, 1F
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