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
(1) ∵ the line L intersects with the parabola x ^ 2 = 4Y at two points, and∵ the line L has a slope, so that its slope is K. from the parabola equation x ^ 2 = 4Y, the coordinates of its focus f are (1,0), ∵ the equation of the line L is y = KX + 1. ∵ A and B are on the line y = KX + 1, ∵ the coordinates of a and B can be respectively (m, KM + 1), (n, K
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- 1. 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
- 2. 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
- 3. 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)=______ .
- 4. 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
- 5. (1)2x-3y=5 x+2y=1 (2)u-9.8t-1=0 u/3-1=3t (3)x-5y=16 5y-5x=4 (4)2x-3y=5 x+y=5/3 (5)21x+23y=243 23x+21y=241 Find x, Y. quick, I'll make it clear. (1)2x-3y=5 x+2y=1 (2)u-9.8t-1=0 u/3-1=3t (3)x-5y=16 5y-5x=4 (4)2x-3y=5 x+y=5/3 (5)21x+23y=243 23x+21y=241
- 6. Given the function f (x) = x & # 178; + 2aX + 2, X ∈ [- 5,5] Use a to represent the maximum value of function f (x) in the interval [- 5,5],
- 7. If the function f (x) = ax ^ 4 + BX ^ 2 + C is known and the tangent equation at x = 1 is y = X-2, then the analytic expression of F (x) is given
- 8. The function f (x) = - x ^ 3 + ax ^ 2 + BX + C is known. The tangent equation at P (1, - 2) on the image is y = - 3 + 1 If the function f (x) increases monotonically in the interval [- 2,0], the value range of real number B is obtained
- 9. The tangent equation y = X-1 at the beginning of point (1, f (1)) for the image with function f (x) = ax + BX + C (a > 0) is known (I) B, C are expressed by A; (II) if f (x) > LNX is constant on [1, ∞], find the value range of A; (III) proof: 1 + 1 / 2 + 1 / 3 +. + 1 / N ≥ ln (n + 1) + n / 2 (n + 1) (n ≥ 1)
- 10. Let a be a constant and a > 1,0 < x ≤ 2 π, then f (x) = cos ^ 2x Let a be a constant, and a > 1, 0 be less than or equal to x, 2 be less than or equal to, and find the maximum value of the function f (x) = cos square x + 2asinx-1
- 11. 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
- 12. 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
- 13. Tangent equation of a point P (2.8 / 3) passing through the square of the curve y = 1 / 3x
- 14. If the line 3x-y + 4 = 0 and 6x-2y-1 = 0 are two tangents of a circle, the area of the circle is
- 15. 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
- 16. Find the tangent equation and normal plane equation of the curve X = t, y = T ^ 2, z = T ^ 3 at t = 2
- 17. 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
- 18. 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
- 19. Finding tangent and normal plane equation of curve y ^ 2 = 2mx Z ^ 2 = M-X at point (x.y.z.)
- 20. 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)