The tangent equation of curve y = x ^ 3 + 3x-8 at x = 2 is We need to explain it in detail

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• 1. Find the tangent and normal plane equation of the curve X ^ 2-y ^ 2 = 3 and x ^ 2 + y ^ 2-z ^ 2 = 4 at the point (- 2, - 1,1)
• 2. Finding tangent and normal plane equation of curve y ^ 2 = 2mx Z ^ 2 = M-X at point (x.y.z.)
• 3. Finding the tangent equation of curve X ^ 2 + y ^ 2 + Z ^ 2 = 2, x + y + Z = 0 at point (1,0. - 1) is a normal plane equation
• 4. Find a point on the curve Γ: x = t, y = T ^ 2, z = T ^ 3 so that the tangent of the point is parallel to the plane π: x + 2Y + Z = 4
• 5. Find the tangent equation and normal plane equation of the curve X = t, y = T ^ 2, z = T ^ 3 at t = 2
• 6. Among all the tangents of the curve X = t, y = - T ^ 2, z = T ^ 3, how many are parallel to the plane x + 2Y + Z = 4? Hello teacher, my solution is like this: 1 find the tangent vector (1, - 2T, 3T ^ 2) 2 orthogonal to the normal vector of the plane (1,2,1) can get 1-4t + 3T ^ 2 = 0, the solution is t = 1 / 3, t = 1, so there are two, but what I don't understand is why the answer adds a restriction, that is, after finding t, bring in the curve to get the tangent point P1 (1 / 3, - 1 / 9, 1 / 27), P2 (1, - 1, 1) Because the tangent point is not on the plane, the final answer is two. What I want to ask is that even if the tangent point is on the plane, the tangent vector within the plane is parallel to the plane. Why should it be regarded as a limiting condition
• 7. If the line 3x-y + 4 = 0 and 6x-2y-1 = 0 are two tangents of a circle, the area of the circle is
• 8. Tangent equation of a point P (2.8 / 3) passing through the square of the curve y = 1 / 3x
• 9. Through two points (- 1,1 / 2) on the parabola x ^ 2 = 2Y, B (2,2) respectively make the tangent line of the parabola, and the two tangent lines intersect at point M Verification ∠ BAM = ∠ BMA
• 10. It is known that the straight line L: y = KX + m intersecting parabola C: x ^ 2 = 4Y is tangent to two different points a and B respectively Let two tangents intersect at point M. if M (2, - 1), the equation of line L is obtained
• 11. 1. A straight line is cut by two straight lines L1: 4x + y + 6 = 0, L2: 3x-5y-6 = 0, and the midpoint of the line segment is P point. When the coordinate of P point is (0,0), find the straight line 1. A straight line is cut by two straight lines L1: 4x + y + 6 = 0, L2: 3x-5y-6 = 0, and the midpoint of the line segment is P point. When the coordinate of P point is (0,0), the linear equation is solved 2. If the line L and two lines y = 1, x-y-7 = 0 intersect at P and Q respectively, the middle point of the line PQ is (1, - 1) the equation for solving L
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• 20. For the point (x0, Y0, Z0), t tends to 0, and the function f () satisfies f（x0+t,y,z)=f(x0,y0,z0)*P(y-y0,z-z0）； Where p () is a two-dimensional normal distribution function related to y-y0 and z-z0, Given the initial value of F (x0, Y0, Z0), I want to find the value of any f (x1, y, z) at x = x1, as long as I think about it If you can do it, I'll give you my share The fairy on the first floor f(x0,y0,z0)*P(y-y0,z-z0）-f（x0,y,z) =F (x0, Y0, Z0) (P (y-y0, z-z0) - 1) is not very right, I'm changing f (x0 + T, y, z) = f (x0, Y0, Z0) * P (y-y0, z-z0) × exp (- at); a is a known constant