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

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 $x y = k$
$\to 16 x^{2} + y^{2} = c$
Differentiating w.r.t. $x$
$32 x + 2 y \frac{d y}{d x} = 0$
$\Rightarrow \frac{d y}{d x} = m_{1} = - \frac{16 x}{y}$
and $y^{16} = k x$
Differentiating w.r.t. $x$
$16 y^{15} \frac{d y}{d x} = k$
$\Rightarrow \frac{d y}{d x} = m_{2} = \frac{k}{16 y^{15}}$
$m_{1} m_{2} = - \frac{16 x}{y} \times \frac{k}{16 y^{15}} = - \frac{x}{y^{16}} \times k$
$\Rightarrow m_{1} m_{2} = - \frac{x}{y^{16}} \times \frac{y^{16}}{x} = - 1$

$\to y = x + c e^{- x}$
$\Rightarrow \frac{d y}{d x} = 1 - c e^{- x} = 1 - (y - x) = - (y - x - 1)$
$\Rightarrow \frac{d y}{d x} = m_{1} = - (y - x - 1)$
and $x + 2 = y + k e^{- y}$
$\Rightarrow \frac{d y}{d x} - k \frac{d y}{d x} \cdot e^{- y} = 1$
$\Rightarrow \frac{d y}{d x} [1 - k e^{- y}] = 1$
$\Rightarrow \frac{d y}{d x} = m_{2} = \frac{1}{1 - k e^{- y}} = \frac{1}{y - x - 1}$
$\Rightarrow m_{1} m_{2} = - 1$

$\to y = c x^{2}$
$\Rightarrow \frac{d y}{d x} = 2 c x = 2 x \cdot \frac{y}{x^{2}} = \frac{2 y}{x}$
$\Rightarrow \frac{d y}{d x} = m_{1} = \frac{2 y}{x}$
and $x^{2} + 2 y^{2} = k$
$\Rightarrow 2 x + 4 y \frac{d y}{d x} = 0$
$\Rightarrow \frac{d y}{d x} = m_{2} = - \frac{x}{2 y}$
$\Rightarrow m_{1} m_{2} = - 1$

$\to x^{2} - y^{2} = c$
$\Rightarrow 2 x - 2 y \frac{d y}{d x} = 0$
$\Rightarrow \frac{d y}{d x} = m_{1} = \frac{x}{y}$
and $x y = k$
$\Rightarrow x \frac{d y}{d x} + y = 0$
$\Rightarrow \frac{d y}{d x} = m_{2} = - \frac{y}{x}$
$\Rightarrow m_{1} m_{2} = - 1$

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