many thanks get the planes tangent on after areas at indicated points: x2 + 2y2 + 3xz = 10, at point (1, 2. 1/3 ) y2 – x2 = 3. at point (1, 2, 8) xyz = 1, at point (1, 1. 1)

Think about the after equation of area provided point: 1,2, allow ,f(x,y,z)=x” + 2y’ +3xz-10 The gradient of area is, är =2x+0+32(1)-0 s, -(x+2y+30) ду 0 2 (2y)+0-0 左-(x2 +2y2 +3xz-10) Oz =0+0+ 3x(1)-0 =3x (EN ) = 3(1)

-2(x-1) +4(y-2)-0(z-8)=0 -2x+ 2+4y-8 0 2x-4y +6 = 0 think about the after equation of area nz = 1 provided point: (1.1,1) Allow f(x,y,z) =nz-1 The gradient of area is, (nz-1 左 =yz (x)–(1) ox ax = yz (A)(LU)-1 Oz Ozz =xy-1-0 (ALLI)-1-1

The vector letter is normal on provided area is, n (Vfan (LLl) 1,1,1 The equation of airplane is, x1-1+1(z-1) 0 xtytz-30

小匙2-/-3) oz 3 Oz oz oz = 0-0-0 0 2 12,8 The vector letter is normal on provided area is (12.8) = (- The equation of airplane is, ) 2,4,0