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For the curve `y=4x^3-2x^5,`find all the points at which the tangents pass through the origin. |
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Answer» The equation of the given curve is `y = 4x^(3) - 2x^(5)`..(i) On differentiating (i) w.r.t. x, we get `(dy)/(dx) = 12 x^(2) - 10x^(4)` Let the required point be `P (x, y)` The equation of tangent to be given curve at P(x, y) is given by `Y - y = (12x^(2) - 10x^(4)) (X - x)`...(ii) If it passes through (0, 0) then, we have `y = (12x^(2) - 10x^(4)) x` `rArr 4x^(3) - 2x^(5) = 12x^(3) - 10x^(5)` [using (i)] `8x^(3) - 8x^(5) = 0 rArr 8x^(3) (1-x^(2)) = 0` `rArr x = 0 or x = 1 or x = -1` Putting x = 0 in (i), we get y = 0 Putting x = 1 in (i), we get y = 2 Putting `x = -1` in (i), we get `y = -2` So, the required points are (0, 0) , (1, 2) and `(-1, -2)` |
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