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Find the curves for which the length of normal is equal to the radius vector. |
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Answer» The length of normal ` |y| sqrt(y+((dy)/(dx))^(2))` From the hypothesis of the problem `|y| sqrt(1+((dy)/(dx))^(2))` = radius vector `= r = sqrt(x^(2) + y^(2))` `rArr y^(2){1+ ((dy)/(dx))^(2)} = r^(2) = x^(2) + y^(2)` `rArr y^(2) + y^(2)((dy)/(dx))^(2) = x^(2) + y^(2) " " rArr y^(2)((dy)/(dx))^(2) = x^(2)` `rArr +- y(dy)/(dx) = x " " rArr +- ydy = xdx` `rArr +- y^(2) = x^(2) - k^(2) " " :. x^(2) +- y^(2) = k^(2)` This represents a circle or equilateral hyperbola according as we take + or - sign. |
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