Question

Which of the following pairs is/are orthogonal?
($c$ and $k$ are arbitrary constants.)

• A. $16x^2+y^2=c$ and $y^{16}=kx$
• B. $y=x+c{e}^{-x}$ and $x+2=y+k{e}^{-y}$
• C. $y=c{x}^2$ and $x^2+2y^2=k$
• D. $x^2-y^2=c$ and $xy=k$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $16x^2+y^2=c$ and $y^{16}=kx$
B $y=x+c{e}^{-x}$ and $x+2=y+k{e}^{-y}$
C $y=c{x}^2$ and $x^2+2y^2=k$
D $x^2-y^2=c$ and $xy=k$

$\to 16x^2+y^2=c$
Differentiating w.r.t. $x$
$32x+2y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=m_1=-\frac{16x}{y}$
and $y^{16}=kx$
Differentiating w.r.t. $x$
$16y^{15}\frac{dy}{dx}=k$
$\Rightarrow \frac{dy}{dx}=m_2=\frac{k}{16y^{15}}$
$m_1{m}_2=-\frac{16x}{y}\times \frac{k}{16y^{15}}=-\frac{x}{y^{16}}\times k$
$\Rightarrow m_1{m}_2=-\frac{x}{y^{16}}\times \frac{y^{16}}{x}=-1$

$\to y=x+c{e}^{-x}$
$\Rightarrow \frac{dy}{dx}=1-c{e}^{-x}=1-(y-x)=-(y-x-1)$
$\Rightarrow \frac{dy}{dx}=m_1=-(y-x-1)$
and $x+2=y+k{e}^{-y}$
$\Rightarrow \frac{dy}{dx}-k\frac{dy}{dx}\cdot e^{-y}=1$
$\Rightarrow \frac{dy}{dx}[1-k{e}^{-y}]=1$
$\Rightarrow \frac{dy}{dx}=m_2=\frac{1}{1-k{e}^{-y}}=\frac{1}{y-x-1}$
$\Rightarrow m_1{m}_2=-1$

$\to y=c{x}^2$
$\Rightarrow \frac{dy}{dx}=2cx=2x\cdot \frac{y}{x^2}=\frac{2y}{x}$
$\Rightarrow \frac{dy}{dx}=m_1=\frac{2y}{x}$
and $x^2+2y^2=k$
$\Rightarrow 2x+4y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=m_2=-\frac{x}{2y}$
$\Rightarrow m_1{m}_2=-1$

$\to x^2-y^2=c$
$\Rightarrow 2x-2y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=m_1=\frac{x}{y}$
and $xy=k$
$\Rightarrow x\frac{dy}{dx}+y=0$
$\Rightarrow \frac{dy}{dx}=m_2=-\frac{y}{x}$
$\Rightarrow m_1{m}_2=-1$