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A tangent at a point P on the curve cuts the x-axis at A and B is the foot of perpendicular from P on the x axis. If the midpoint of AB is fixed at `(alpha,0)` for any point P, find the differential equation and hence find the curve. |
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Answer» here, ` y=f(x)` `Y - y = dy/dx*(X-x)` putting Y=0`-y = dy/dx*(X-x)` `X= x- y*dy/dx` C is the mid point of AB let coorinates of C as `(alpha,0)` `C(alpha,0) = ((x+x - y*(dy/dx))/2 , 0)` `2*alpha = 2*x - y*dx/dy` `y*dx/dy = 2*(x- alpha)` this is differential eqn integrating both sides `int dx/(2*(x-alpha)) = int dy/y` `1/2 * ln(x-alpha) = y + c` `ln(x- alpha) = 2*y +2*c` `ln(x-alpha) = 2*y + k` `x-alpha = e^(2*y+k)` `x= alpha + e^(2y+k)` |
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