#### To determine

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

#### Explanation

**Given:**

The curves y=cx2 and x2+2y2=k.

**Derivative rules:**

*Chain rule*: dydx=dydu⋅dudx

**Proof:**

Obtain the slope of the curves y=cx2.

ddx(y)=ddx(cx2)dydx=cddx(x2)=2cx

Thus the slope of the tangent to the curve y=cx2 is m1=2cx.

Obtain the slope of the curves x2+y2=k.

ddx(x2+y2)=ddx(k)ddx(x2)+2ddx(y2)=02x+4ydydx=0dydx=−x2y

Thus, the slope of the equation x2+y2=k is m2=−x2y.

The product slope is computed as follows,

m1m2=2cx(−x2y)=−2cx22y=−2y2y [Qy=cx2]=−1

If c≠0 then the family of the curves orthogonal trajectories to another family of the curves.

If c=0, then

y=(0)x2=0

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

x2+(0)2=kx2=kx=±k

The line y=0 is horizontal line and it intersect x2+2y2=k at the points (−k,0) and (k,0). That is, the curve x2+2y2=k has vertical tangent at the points (−k,0) and (k,0).

Therefore, the curves y=cx2 and x2+y2=k are orthogonal trajectories to each other is proved.

**Graph:**

The family of the curves y=cx2 and x2+y2=k as shown below in Figure 1.

From the graph, it is observed that the family of the curves y=cx2 and x2+y2=k are orthogonal trajectories to each other.