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Find the equation of the curve passing through the point, (5,4) if the sum of reciprocal of the intercepts of the normal drawn at any point P(x,y) on it is 1.

Text Solution

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The equation of the normal at (x,y) is
`(X-x)+(Y-y)(dy)/(dx)=0`
or `X/(x+y(dy)/(dx))+Y/(x+ydy//dx)/(dy//dx)=1`
x-intercept OA `=x+y(dy)/(dx)`,
x-intercept OB `=(x+y(dy)(dx))/((dy)/(dx))`
Given, `1/(OA) + 1/(OB) =1`
or `1+(dy)/(dx) = x+y(dy)/(dx)`
or `(y-1)dy+(x-1)dx=0`
Integrating, we get
`(y-1)^(2)+(x-1)^(2)=c`
Since, the curve passes through `(5,4), c=25`.
Hence, the curve is `(x-1)^(2)+(y-1)^(2)=25`.
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