InterviewSolution
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InterviewSolution
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A curve passes through the point (-2, 1) and at any point (x, y) of the curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4, -3). Find the equation of the curve. |
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Answer» Correct Answer - `y+3=(x+4)^(2)` `(dy)/(dx) = (2(y+3))/((x+4))rArr int (1)/((y+3))dy =2 int (1)/((x+4))dx` `therefore log |y+3|=2log|x+4|+log|C_(1)| rArr ((y+3))/((x+4)^(2)) =pm C_(1)=C.` Putting x = -2 and y = 1, we get C = 1. |
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