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The differential equation representing all possible curves that cut each member of the family of circles `x^(2)+y^(2)-2Cx=0` (C is a parameter) at right angle, isA. `(dy)/(dx)=(2xy)/(x^(2)+y^(2))`B. `(dy)/(dx)=(2xy)/(x^(2)-y^(2))`C. `(dy)/(dx)=(x^(2)+y^(2))/(2xy)`D. `(dy)/(dx)=(x^(2)-y^(2))/(2xy)` |
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Answer» Correct Answer - B Here, we have to find the orthogonal trajectories of the family of circles `x^(2)+y^(2)-2Cx=0" …(i)"` Differentiating (i) w.r.t. x, we get `2x+2y(dy)/(dx)-2C=0rArrC=x+y(dy)/(dx)" ...(ii)"` From (i) and (ii), we obtain `x^(2)+y^(2)-2x(x+y(dy)/(dx))=0" [By eliminating C]"` `rArr" "y^(2)-x^(2)-2xy(dy)/(dx)=0rArry^(2)-x^(2)=2xy(dy)/(dx)" ...(iii)"` This is the differential equation representing the given family of circles. To find the differential equation of the orthogonal trajectories, we replace `(dy)/(dx)by-(dx)/(dy)` in equation (iii). Thus, the differential equation representing the orthogonal trajectories is `y^(2)-x^(2)=-2xy(dx)/(dy)or, (dy)/(dx)=-(2xy)/(x^(2)-y^(2))` |
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