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Find the coordinates of the points on the curve `y=x^2+3x+4,`the tangents at which pass through the origin. |
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Answer» Let the required point be `(x_(1) , y_(1))` Now, `y = x^(2) + 3x + 4 rArr (dy)/(dx) = 2x + 3 rArr ((dy)/(dx))_((x_(1)","y_(1))) = (2x_(1) + 3)` So, the equation of the tangent at `(x_(1), y_(1))` is `(y - y_(1))/(x - x_(1)) = (2x_(1) + 3)` Since the tangent passes through the origin, we have `(-y_(1))/(-x_(1)) = (2x_(1) + 3)` `rArr y_(1) = (2x_(1)^(2) + 3x_(1))` ...(i) But, `(x_(1), y_(1))` lies on the given curve, So, `y_(1) = x_(1)^(2) + 3x_(1) + 4`...(ii) From (i) and (ii), we get `2x_(1)^(2) + 3x_(1) = x_(1)^(2) + 3x_(1) + 4 or x_(1)^(2) = 4 or x_(1) = +-2` Now, `x_(1) = 2 rArr y_(1) = (2^(2) + 3 xx 2 + 4) = 14` And, `x_(1) = -2 rArr y_(1) = [(-2)^(2) + 3 xx (-2) + 4] = 2` Hence, the required points are (2, 14) and `(-2, 1)` |
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