Question

53-56 Two curves are orthogonal if their tangent lines are perpendicular at each point of intersection. Show that given families of curves are orthogonal trajectories of each other; that is, every curve in one family is orthogonal to every curve in the other family. Sketch both families of curves on the same axes.

55. \(y = c{x^{\bf{2}}}\), \({x^{\bf{2}}} + {\bf{2}}{y^{\bf{2}}} = k\)

Short answer

The sketch of curves is shown below:

Step-by-step solution

Differentiate the equations of the given curves

Differentiate the equation \(y = c{x^2}\)with respect to \(x\)as follows:

\(\begin{aligned}y' &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {c{x^2}} \right)\\ &= 2cx\end{aligned}\)

Differentiate the equation \({x^2} + 2{y^2} = k\)with respect to \(x\)as follows:

\(\begin{aligned}2x + 4yy' &= 0\\2yy' &= - x\\y' &= - \frac{x}{{2y}}\end{aligned}\)

Simplify the equations of \(y'\)

Substitute \(c{x^2}\) for \(y\) in the equation \(y' = - \frac{x}{{2y}}\).

\(\begin{array}{c}y' = - \frac{x}{{2\left( {c{x^2}} \right)}}\\ = - \frac{1}{{2cx}}\end{array}\)

So, the given curves are orthogonal to each other, if \(c \ne 0\). When \(c = 0\), the curve \(y = c{x^2}\) will be \(y = 0\), which is a horizontal line. The intersection point of the line \(y = 0\) and \({x^2} + 2{y^2} = k\) is \(\left( { \pm \sqrt k ,0} \right)\)

Sketch the orthogonal trajectory

So, the curve \(y = c{x^2}\) and \({x^2} + 2{y^2} = k\) intersects orthogonally.

The figure below represents the orthogonal trajectory of curves.