Question

The orthogonal trajectories for $y=C{e}^{-x}$ represent

• A. exponential curve
• B. family of ellipse
• C. family of circles
• D. family of parabolas

Answer 1

The correct option is D family of parabolas

$y{e}^x=C$
Differentiating, $y^{\prime }{e}^x+e^{x}y=0\Rightarrow y^{\prime }+y=0$.
Replace $y^{\prime }$
by $-\frac{dx}{dy}$,
we have $1-y\frac{dy}{dx}=0\Rightarrow ydy=dx\Rightarrow \frac{y^2}{2}=x+C$
which represent a family of parabolas.