Find the tangent equation and normal plane equation of the curve X = t, y = T ^ 2, z = T ^ 3 at t = 2

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• 1. 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
• 2. If the line 3x-y + 4 = 0 and 6x-2y-1 = 0 are two tangents of a circle, the area of the circle is
• 3. Tangent equation of a point P (2.8 / 3) passing through the square of the curve y = 1 / 3x
• 4. 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
• 5. 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
• 6. X ^ 2 = 4Y, the tangent of line L passing through a and B is L1 and L2 (1) Verification of L1 vertical L2 (2) It is proved that the focus of L1 and L2 is on the directrix
• 7. The parabola x 2 = 4Y P is a tangent line of circle x 2 + (y + 1) 2 = 1 crossed by a moving point on the parabola passing through point P, y = - 2 at two points ab. when Pb is just tangent to the parabola and point P, the area of △ PAB is calculated
• 8. Given that the tangent equation of the image of the function f (x) = (ax-6) / (x ^ 2 + b) at the point m (- 1, f (- 1)) is x + 2Y + 5 = 0, find the analytic expression graph of the function y = f (x) First, we substitute the point m x = - 1 into the tangent equation and get y = - 2 So the point (- 1, - 2) is the tangent point of F (x) If f (x) is derived, f '(x) = (- ax ^ 2 + 12x + AB) / (x ^ 2 + b) & # 178; So f '(- 1) = (- A-12 + AB) / (1 + b) & # 178; = - 1 / 2 (slope of tangent) ① And f (- 1) = (- a-6) / (B + 1) = - 2 That is, a = 2b-4 From (1) to (2) A = - 6, B = - 1 or a = 2, B = 3 And because (x ^ 2 + b) is the denominator, it is not zero, so B = - 1 is omitted a=2 b=3 Where f '(- 1) = (- A-12 + AB) / (1 + b) &# 178; = - 1 / 2 is based on what? Why is the slope obtained by derivation
• 9. Given that the tangent equation of curve y = f (x) at point (1, f (1)) is x-2y + 4 = 0, then f (1) + F '(1)=______ ．
• 10. To compare the size of two numbers a and B, sometimes we can solve it by the size of A-B and 0: if a = 5x + 2Y, B = 2x = 5Y, and X is less than y, try to compare the size of a and B
• 11. 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
• 12. 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
• 13. Finding tangent and normal plane equation of curve y ^ 2 = 2mx Z ^ 2 = M-X at point (x.y.z.)
• 14. 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)
• 15. The tangent equation of curve y = x ^ 3 + 3x-8 at x = 2 is We need to explain it in detail
• 16. 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
• 17. Find the special solution of y = y '= 0 when y' '+ (y)' ^ 2 = 1 and x = 0
• 18. Finding the special solution of differential equation y '- 2Y / (1-x ^ 2) = x + 1, x = 0, y = 0
• 19. The solution y =? Of the differential equation y '' + 2Y '+ y = 0?
• 20. Finding the special solution of differential equation y '- y = cosx, x = 0, y = 0