Question

53–56:Orthogonal Trajectories: Two curves are orthogonal if

their tangent lines are perpendicular at each point of intersection.Show that the given families of curves are orthogonal trajectoriesof each other; that is, every curve in one family is orthogonal toevery curve in the other family. Sketch both families of curves onthe same axes.

56. \(y = a{x^3},{\rm{ }}{x^2} + 3{y^2} = b\)

Short answer

It is proved that given families of curves are orthogonal trajectories of each other.

Step-by-step solution

Use Implicit differentiation

Differentiate both sides of the equation with respect to \(x\).

\(\begin{array}{c}y = a{x^3}\\\frac{d}{{dx}}\left( y \right) = \frac{d}{{dx}}\left( {a{x^3}} \right)\\y' = 3a{x^2}\end{array}\)

Use Implicit differentiation

Differentiate both sides of the equation with respect to \(x\).

\(\begin{aligned}{x^2} + 3{y^2} &= b\\2x + 6yy' &= 0\\6yy' &= - 2x\\y' &= - \frac{{2x}}{{6y}}\\ &= - \frac{x}{{3y}}\end{aligned}\)

Substitute the values

Substitute \(y = a{x^3}\) into the equation \(y' = - \frac{x}{{3y}}\).

\(\begin{aligned}{c}y' &= - \frac{x}{{3y}}\\ &= - \frac{x}{{3\left( {a{x^3}} \right)}}\\ &= - \frac{1}{{3a{x^2}}}\end{aligned}\)

This implies that both the curves are orthogonal when \(a \ne 0\). If \(a = 0\), as the ellipse has vertical asymptotes at \(\left( {\sqrt b ,0} \right)\) and \(\left( { - \sqrt b ,0} \right)\)then both the curves intersect orthogonally at \(\left( { \pm \sqrt b ,0} \right)\).

Draw the curve of both the functions

Represent the curve of parabola and ellipse. The graph is shown below.