Question

What is the orthogonal trajectory for the family of curves $a{x}^2+b{y}^2=k^2$, where $a$ and $b$ are fixed constants?

• A. $y^{\frac{b}{2}}=c{x}^{\frac{a}{2}}$
• B. $y^{a-b}=c{x}^{a-b}$
• C. $y^b=c{x}^a$
• D. $y^a=c{x}^b$

Answer 1

The correct option is D $y^a=c{x}^b$

We have, $a{x}^2+b{y}^2=k^2\left(1\right)$
Differentiating w.r.t. $x$, we get
$2ax+2by\frac{dy}{dx}=0⟹\frac{dy}{dx}=-\frac{ax}{by}\left(2\right)$

For finding the orthogonal trajectory, we need to
Replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$ in (2),
we get,

$-\frac{dx}{dy}=-\frac{ax}{by}$
$⟹\frac{a}{y}dy=\frac{b}{x}dx$
Integrating the above equation, we get
$a\log y=b\log x+\log c$$⟹y^a=c{x}^b$

Thus this is the required orthogonal trajectory to the given curve.