maths & stat (equ.of tangent)－快遞百科

標題:

maths & stat (equ.of tangent)

發問:

Let C1: x^2 y^2=5 C2: xy=6(a) Find the points of intersection of the curves. (b) At the point of intersection K in the third quadrant, (i) find the equation of the tangent to the curve C1,(ii) find the equation of the tangent to the curve C2,(iii) show that the tangents in (i) and (ii) are prependicular... 顯示更多 Let C1: x^2 y^2=5 C2: xy=6 (a) Find the points of intersection of the curves. (b) At the point of intersection K in the third quadrant, (i) find the equation of the tangent to the curve C1, (ii) find the equation of the tangent to the curve C2, (iii) show that the tangents in (i) and (ii) are prependicular to each other. 我已經計左(a)同 (b)(i) , 但係計唔到(b)(ii) ans: (a) the points of intersection of the curves are (3,2) , (-3,-2) (b)(i) the equation of the tangent to the curve C1 is 3x-2y+5=0 睇番本書(b)(ii)既ans 係2x+3y+12=0 我想知個step呀!! 唔該哂大家....

最佳解答:

(b)(ii) Let the tangent be y = mx + c Since (-3,-2) is on the line, -2 = -3m + c c = 3m - 2 Therefore the line is y = mx + 3m - 2 Sub y = mx + 3m - 2 into C2 x(mx + 3m - 2) = 6 mx^2 + (3m-2)x - 6 = 0 Discriminant (3m-2)^2 + 24m = 0 9m^2 - 12m + 4 + 24m = 0 9m^2 + 12m + 4 = 0 (3m + 2)^2 = 0 m = -2/3 Therefore equation of tangent is y = (-2/3)x + (3)(-2/3) - 2 y = (-2/3)x - 4 3y = -2x - 12 2x + 3y + 12 = 0 (iii) Slope of tangent in (i) is 3/2 Slope of tangent in (ii) is -2/3 Product of tangent = (3/2)(-2/3) = -1 => The tangents are perpendicular