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Find the angle of intersection of the curves `2y^(2) = x^(3) and 32x`. |
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Answer» `2y^(2)= x^(3) …(1)` `y^(2) = 32x` …(2) Solving eqs. (1) and (2) ` 64x = x^(3)` ` rArr x(x^(2)-64) = 0 rArr x=0,8,-8` `:. "At "x=0,y-0` `"At " x=8, y=pm 16` At x =- 8, y is imaginary. Therefore the points of intersection are (0, 0), (8, 16) and (8, -16). From eq.(1) `" "` From eq.(2) `3y^(2) = x^(3)" "y^(2) = 32x` ` rArr 4y (dy)/(dx) = 3x^(2) rArr 2y (dy)/(dx) = 32` `rArr (dy)/(dx) = (3x^(2))/(4y) rArr (dy)/(dx) = (16)/y` At point (8, 16) ` m_(1) = (3(8)^(2))/(4 xx 16) = 3 " "m_(2) = (16)/(16) = 1` ` and tan theta_(1) = (m_(1)-m_(2))/(1+m_(1)m_(2)) = (3-1)/(1+3) = 1/2 ` ` rArr theta_(1) = tan^(-1) 1/2*` At point (8,16) `m_(3)= (3(8)^(2))/(4 xx (-16)) = - 3 " "m_(4) = 16/((-16))=-1` `and tan theta_(2) = (m_(3)-m_(4))/(1+m_(3)m_(4)) = (-3+1)/(1+3) = (-1/2)` `rArr theta_(2) = tan^(-1) (-1/2)*` At point (0, 0), the slopes cannot be determined. |
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