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標題:
A-MATHS
發問:
(1) Find the equation of the two tangents to the curve y=2/ (x-1) which are parallel to the line x+2y-7 =0 (2) The parametric equations of a curve are x= a cos3t , y=a sin3t Find the equation of the tangent to the curve at the point whose parameter is t
最佳解答:
(1) Slope of x + 2y - 7 = 0 is -1/2 For y = 2/(x - 1) dy/dx = -2/(x - 1)2 So, finding for x, we have: dy/dx = -1/2 -2/(x - 1)2 = -1/2 (x - 1)2 = 4 x - 1 = ±2 x = -1 or 3 y = -1 or 1 So tangents at (-1, -1) and (3, 1) are parallel to x + 2y - 7 = 0 Their equations are: (y + 1)/(x + 1) = -1/2 and (y - 1)/(x - 3) = -1/2 2y + 2 = -x - 1 and 2y - 2 = 3 - x x + 2y + 3 = 0 and x + 2y - 5 = 0 (2) Using parametric differentiation: dx/dt = -3a cos2 t sin t dy/dt = 3a sin2 t cos t dy/dx = (dy/dt)/(dx/dt) = - tan t So equation of tangent is given by: (y - a sin3 t)/(x - a cos3 t) = - tan t cos t (y - a sin3 t) = - sin t (x - a cos3 t) y cos t - a sin3 t cos t = -x sin t + a cos3 t sin t x sin t + y cos t - a sin3 t cos t - a cos3 t sin t = 0 x sin t + y cos t - a sin t cos t (sin2 t + cos2 t) = 0 x sin t + y cos t - a sin t cos t = 0
其他解答:
A-MATHS
發問:
(1) Find the equation of the two tangents to the curve y=2/ (x-1) which are parallel to the line x+2y-7 =0 (2) The parametric equations of a curve are x= a cos3t , y=a sin3t Find the equation of the tangent to the curve at the point whose parameter is t
最佳解答:
(1) Slope of x + 2y - 7 = 0 is -1/2 For y = 2/(x - 1) dy/dx = -2/(x - 1)2 So, finding for x, we have: dy/dx = -1/2 -2/(x - 1)2 = -1/2 (x - 1)2 = 4 x - 1 = ±2 x = -1 or 3 y = -1 or 1 So tangents at (-1, -1) and (3, 1) are parallel to x + 2y - 7 = 0 Their equations are: (y + 1)/(x + 1) = -1/2 and (y - 1)/(x - 3) = -1/2 2y + 2 = -x - 1 and 2y - 2 = 3 - x x + 2y + 3 = 0 and x + 2y - 5 = 0 (2) Using parametric differentiation: dx/dt = -3a cos2 t sin t dy/dt = 3a sin2 t cos t dy/dx = (dy/dt)/(dx/dt) = - tan t So equation of tangent is given by: (y - a sin3 t)/(x - a cos3 t) = - tan t cos t (y - a sin3 t) = - sin t (x - a cos3 t) y cos t - a sin3 t cos t = -x sin t + a cos3 t sin t x sin t + y cos t - a sin3 t cos t - a cos3 t sin t = 0 x sin t + y cos t - a sin t cos t (sin2 t + cos2 t) = 0 x sin t + y cos t - a sin t cos t = 0
其他解答:
此文章來自奇摩知識+如有不便請留言告知
(1) slope of x+2y-7=0 is -1/2. Hence slope of the required tangents is -1/2. y=2/(x-1) dy/dx = -2/(x-1)^2 Set dy/dx=-1/2 -2/(x-1)^2 = -1/2 (x-1)^2 = 4 x-1 = 2 or -2 x = 3 or -1. So the points of touch are (3,1) and (-1,-1), and the equations of the tangents are: y-1 = (-1/2)(x-3) i.e. x+2y-5=0 and y-(-1) = (-1/2)(x-(-1)) i.e. x+2y+3=0. (2) dx/dt = -3a cos^2 t sin t dy/dt = 3a sin^2 t cos t So dy/dx = (dy/dt)/(dx/dt) = -sint/cost = -tan t at the point with parameter t.
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