Question

find the curve for which the length of the normal in any of its points is proportional to the square of the ordinate of the point. take the coefficient of proportionality equal to k

Answer 1

You are describing a circle. To see why, consider the unit circle: An arbitrary point (x,y) and it's position vector <x,y> are shown. The vector <x,y> is normal to the curve at any point (x,y):It's length is \[\left| r \right|=(x^2+y^2)^\frac{1}{2}\]which is the square of the ordinate point. Now if we don't restrict ourselves to the unit circle but pick any arbitrary circle, we have:\[r^2=k(x^2+y^2)\]The proportionality constant affects the length of the normal vector, which is proportional to the square of the ordinate point. If we pick k=9 for example at the point (x,y)=(1,1), we get a vector at point (1,1) which is normal to the curve there but has length=3. If we keep k=9 and look at the point (2,0) we have the vector <18,0>, which is normal to the curve at point (2,0) with length=18. Hope this helps!