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Find the orthogonal trajectory of `y^2=4a x`(a being the parameter). |
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Answer» `y^(2)=4ax`…………..(1) `therefore 2y(dy)/(dx)=4a`……………………..(2) Eliminating a from equation (1) and (2), we get `y^(2)=2y(dy)/(dx)x` Replacing `(dy)/(dx)dy-(dx)/(dy)`, we get `y=2(-(dx)/(dy))x` `2xdx+ydy=0` Integrating both sides, we get `x^(2)+y^(2)/(2)=c` `2x^(2)+y^(2)=2c` Which is the required orthogonal trajectory. |
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