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At what point on the curve `x^(3) - 8a^(2) y= 0` the slope of the normal is `-2//3` ?A. (a,a)B. (2a,-a)C. (2a,a)D. None of these |
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Answer» Correct Answer - C Given curve is `x^(2) - 8a^(2)y = 0` On differentiating w.r.t . X, we get ` 3x^(2) - 8a^(2)(dy)/(dx) = 0 rArr (dy)/(dx) = (3x^(2))/(8a^(2))` `therefore` Slope of the normal ` =-(1)/(((dy)/(dx))) = -(1)/((3x^(2))/(8a^(2))) = - (8a^(2))/(3x^(2))` Given ` - (8a^(2))/(3x^(2)) = - (2)/(3)` `rArr " "x^(2) = 4a^(2) rArr x = pm 2a` At `" "x = pm 2a, " "y = pm a` `therefore " "(x,y) = (2a,a)` |
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