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

16 x^{2} + y^{2} = c

and

y^{16} = k x

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y = x + c e^{- x}

and

x + 2 = y + k e^{- y}

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y = c x^{2}

and

x^{2} + 2 y^{2} = k

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x^{2} - y^{2} = c

and

x y = k

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Solution

The correct option is D $x^{2} - y^{2} = c$ and $x y = k$
$16 x^{2} + y^{2} = c$ and $y^{16} = k x$
$\Rightarrow 32 x + 2 y y^{'} = 0 \Rightarrow y_{1}^{'} = - \frac{16 x}{y} y^{16} = k x 16 y^{15} y_{2}^{'} = k \Rightarrow y_{2}^{'} = \frac{k}{16 y^{15}} \Rightarrow y_{1}^{'} y_{2}^{'} = - \frac{16 x}{y} \times \frac{k}{16 y^{15}} \Rightarrow y_{1}^{'} y_{2}^{'} = - 1$
The curves are orthogonal.

$y = x + c e^{- x}$ and $x + 2 = y + k e^{- y}$
$y_{1}^{'} = 1 - c e^{- x} 1 = y_{2}^{'} - k e^{- y} y_{2}^{'} y_{1}^{'} = 1 - (y - x) = 1 + x - y y_{2}^{'} = \frac{1}{1 - k e^{- y}} \Rightarrow y_{2}^{'} = \frac{1}{1 - (x + 2 - y)} \Rightarrow y_{2}^{'} = \frac{1}{- (1 + x - y)} \Rightarrow y_{1}^{'} y_{2}^{'} = - 1$
The curves are orthogonal.

$y = c x^{2}$ and $x^{2} + 2 y^{2} = k$
$y_{1}^{'} = 2 c x = \frac{2 y}{x} y_{2}^{'} = - \frac{x}{2 y} \Rightarrow y_{1}^{'} y_{2}^{'} = - 1$
The curves are orthogonal.

$x^{2} - y^{2} = c$ and $x y = k$
$y_{1}^{'} = \frac{x}{y} y_{2}^{'} = - \frac{y}{x} \Rightarrow y_{1}^{'} y_{2}^{'} = - 1$
The curves are orthogonal.

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