#### To determine

**To show:** One family of the curves are orthogonal trajectories to the other family.

#### Explanation

**Given:**

The family of the curves y=ax3 and x2+3y2=b.

**Derivative rules:**

*Chain rule*: dydx=dydu⋅dudx.

**Proof:**

Obtain the slope of the curves y=ax3.

ddx(y)=ddx(ax3)dydx=addx(x3)=a(3x2)=3ax2

Thus, the slope of the tangent to the curve y=cx2 is m1=3ax2.

Obtain the slope of the curves x2+3y2=b.

ddx(x2+3y2)=ddx(b)ddx(x2)+3ddx(y2)=02x+6ydydx=0dydx=−x3y

Thus, the slope of the equation x2+3y2=b is m2=−x3y.

The product of slope m1m2 is computed as follows,

m1m2=3ax2(−x3y)=−3ax33y=−3y3y [Qy=ax3]=−1

If a≠0, then the family of the curves the curves y=ax3 and x2+3y2=b are orthogonal trajectories .

If a=0, then

y=(0)x3=0

Substitute the value y=0 in the equation x2+3y2=b.

x2+(0)2=bx2=bx=±b

The line y=0 is horizontal line and it is intersect on the ellipse orthogonally at the point (−b,0) and (b,0). That is, the ellipse has vertical tangent at (−b,0) and (b,0).

Therefore, the family of the curves y=ax3 and x2+3y2=b are orthogonal trajectories to each other is proved.

**Graph:**

The family of the curves y=ax3 and x2+3y2=b as shown below in Figure 1.

From Figure 1, it is observed that the family of the curves y=ax3 and x2+3y2=b are orthogonal trajectories.