Question

Find the value of the number a such that the families of curves y=(x+c)−1and y=a(x+k)13 are orthogonal trajectories.

Answer 1

The given curves are

`y_(1)=(x+c)-1` and
`y=a(x+k)+13`

Differentiating both the curves with respect to 'x' we get

`(dy_(1))/(dx)=m_(1)=(d(x+c)-1)/(dx)\\\\=1\\Similarly\\(dy_(2))/(dx)=m_(2)=(d(ax+ak)+13)/(dx)\\\\=a`

Now for orthogonal trajectories we have product of slopes should be -1

`\Rightarrow m_(1)* m_(2)=-1\\\\\\1* a=-1\\\\\therefore a=(-1)/(1)=-1`