In this question, you should consider only points lying in the first quadrant, that is with \(x > 0\) and \(y > 0\).
The equation \(x^2 + y^2 = 2ax\) defines a \emph{family} of curves in the first quadrant, one curve for each positive value of \(a\). A second family of curves in the first quadrant is defined by the equation \(x^2 + y^2 = 2by\), where \(b > 0\).
Differentiate the equation \(x^2 + y^2 = 2ax\) implicitly with respect to \(x\), and hence show that every curve in the first family satisfies the differential equation
\[2xy\frac{\mathrm{d}y}{\mathrm{d}x} = y^2 - x^2.\]
Find similarly a differential equation, independent of \(b\), for the second family of curves.
Hence, or otherwise, show that, at every point with \(y \neq x\) where a curve in the first family meets a curve in the second family, the tangents to the two curves are perpendicular.
A curve in the first family meets a curve in the second family at \((c,\,c)\), where \(c > 0\). Find the equations of the tangents to the two curves at this point. Is it true that where a curve in the first family meets a curve in the second family on the line \(y = x\), the tangents to the two curves are perpendicular?
Given the family of curves in the first quadrant \(y = c\ln x\), where \(c\) takes any non-zero value, find, by solving an appropriate differential equation, a second family of curves with the property that at every point where a curve in the first family meets a curve in the second family, the tangents to the two curves are perpendicular.
A family of curves in the first quadrant is defined by the equation \(y^2 = 4k(x + k)\), where \(k\) takes any non-zero value.
Show that, at every point where one curve in this family meets a second curve in the family, the tangents to the two curves are perpendicular.