Question

Which of the following pair of curves is/are orthogonal?
(where $c$ and $k$ arbitrary constant)

• 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 $xy=k$

$16x^2+y^2=c$ and $y^{16}=kx$
$\Rightarrow 32x+2y{y}^{\prime }=0\Rightarrow y_1^{\prime }=-\frac{16x}{y}{y}^{16}=kx16y^{15}{y}_2^{\prime }=k\Rightarrow y_2^{\prime }=\frac{k}{16y^{15}}\Rightarrow y_1^{\prime }{y}_2^{\prime }=-\frac{16x}{y}\times \frac{k}{16y^{15}}\Rightarrow y_1^{\prime }{y}_2^{\prime }=-1$
The curves are orthogonal.

$y=x+c{e}^{-x}$ and $x+2=y+k{e}^{-y}$
$y_1^{\prime }=1-c{e}^{-x}1=y_2^{\prime }-k{e}^{-y}{y}_2^{\prime }{y}_1^{\prime }=1-(y-x)=1+x-y{y}_2^{\prime }=\frac{1}{1-k{e}^{-y}}\Rightarrow y_2^{\prime }=\frac{1}{1-(x+2-y)}\Rightarrow y_2^{\prime }=\frac{1}{-(1+x-y)}\Rightarrow y_1^{\prime }{y}_2^{\prime }=-1$
The curves are orthogonal.

$y=c{x}^2$ and $x^2+2y^2=k$
$y_1^{\prime }=2cx=\frac{2y}{x}{y}_2^{\prime }=-\frac{x}{2y}\Rightarrow y_1^{\prime }{y}_2^{\prime }=-1$
The curves are orthogonal.

$x^2-y^2=c$ and $xy=k$
$y_1^{\prime }=\frac{x}{y}{y}_2^{\prime }=-\frac{y}{x}\Rightarrow y_1^{\prime }{y}_2^{\prime }=-1$
The curves are orthogonal.