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A normal is drawn at a point P(x , y) of...

A normal is drawn at a point `P(x , y)` of a curve. It meets the x-axis at `Qdot` If `P Q` has constant length `k ,` then show that the differential equation describing such curves is `y(dy)/(dx)=+-sqrt(k^2-y^2)` . Find the equation of such a curve passing through `(0, k)dot`

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The correct Answer is:
`y=cx^(m//n)`
The equation of normal to required curve at P(x,y) is given by `(X-x) + (dy)/(dx)(Y-y)=0`
For point Q, where is this normal meets X-axis, put Y=0. then `X=x+y(dy)/(dx)`
`therefore Q=(x+y(dy)/(dx),0)`

According to question, length of PQ=k.
or `(y(dy)/(dx))^(2)+y^(2)=k^(2)`
or `y(dy)/(dx)=+-sqrt(k^(2)-y^(2))`
Which is the required differential equation of given curve.
Solving this, we get
`int(ydx)/sqrt(k^(2)-y^(2))=int+-dx`
or `-1/22sqrt(k^(2)-y^(2))=+-x+C`
or `-sqrt(k^(2)-y^(2))=+-x+C`
As it passes through (0,k), we get C=0
Thus, equation of curve is `-sqrt(k^(2)-y^(2))=+x`
or `x^(2)+y^(2)=k^(2)`
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