At any point (x, y) of a 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 given that it passes through `(2," "1)` .

Answer

`y=-1/2(x-4)^2+3`

Explanation

We are given that the slope of the tangent is `(dy)/(dx)`

and the slope of the line segment joining the point `(x,y)` and `(4,3)` is
`(y-3)/(x-4)`


Also we are given that the slope of tangent is twice to the slope of the line segment, so


`(dy)/(dx)=2*((y-3)/(x-4))`


`rArr 1/(y-3)dy=2*(1/(x-4))dx`


Now integrating both sides we get


`rArr int 1/(y-3)dy=int 2*(1/(x-4))dx`


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