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The orthogonal trajectories of the family of circles given by `x^2 + y^2 - 2ay = 0`, isA. `x^(2)+y^(2)-2kx=0`B. `x^(2)+y^(2)-2ky=0`C. `x^(2)+y^(2)-2k_(1)x-2k_(2)y=0`D. none of these |
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Answer» Correct Answer - A The equation of the family of circles is `x^(2)+y^(2)-2ay=0" …(i)"` Differentiating w.r.t. x, we get `2x+2y(dy)/(Dx)-2a(dy)/(dx)=0rArra=(x+y(dy)/(dx))/((dy)/(dx))` Putting this value of a in (i), we get `x^(2)+y^(2)-2y{(x+y(dy)/(dx))/((dy)/(dx))}=0rArrx^(2)-y^(2)-2xy(dx)/(dy)=0" ...(ii)"` This is the differential equation of the family of circles givne by (i). The differential equation representing the family of orthogonal trajectories of (i) is obtained by replacing `(dy)/(dx)by-(dx)/(dy)` in (ii). So, the differential equation of the orthogoanl trajectories is `x^(2)-y^(2)+2xy(dy)/(dx)=0` `rArr" "(x^(2)-y^(2))dx+2xydy=0` `rArr" "(xd(y^(2))-y^(2))/(x^(2))=-dx` `rArr" "d((y^(2))/(x))=-dx` `rArr" "(y^(2))/(x)=-x+2krArrx^(2)+y^(2)-2kx=0` This is the required family of orthogonal trajectories. |
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