Which of the following curves cuts the parabola $y^{2} = 4 a x$ at right angle

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

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y = e^{- x / 2 a}

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y = a x

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x^{2} = 4 a y

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Solution

The correct option is C $y = e^{- x / 2 a}$
$y^{2} = 4 a x \Rightarrow 2 y (\frac{d y}{d x}) = 4 a \Rightarrow (\frac{d y}{d x}) = \frac{2 a}{y}$ ...(1)
So, Slope, $m_{1} = \frac{2 a}{y}$

Taking curve $y = e^{- x / 2 a}$
$(\frac{d y}{d x}) = e^{- x / 2 a} (- \frac{1}{2 a}) = - \frac{y}{2 a}$ ...(2)
So, Slope, $m_{2} = - \frac{y}{2 a}$

Both curves cut orthogonaly if
$m_{1} . m_{2} = - 1 \Rightarrow (\frac{2 a}{y}) \times (- \frac{y}{2 a}) = - 1$

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Similar questions

Which one of the following curves cuts the parabola $y^{2} = 4 a x$ at right angles

y^{2} = 4 a x

(a, 2 a)

x^{2} = 4 a y

x^{2} = 4 a y

y^{2} = 4 a x

y^{2} = 4 a x

x^{2} = 4 a y

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

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