Which one of the following curves cuts the pa | vedclass

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(B) For the parabola $y^2 = 4ax$,differentiating with respect to $x$ gives:$2y \frac{dy}{dx} = 4a \implies \left(\frac{dy}{dx}\right)_1 = \frac{2a}{y}

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$y = e^{-x/2a}$

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(B) For the parabola $y^2 = 4ax$,differentiating with respect to $x$ gives:$2y \frac{dy}{dx} = 4a \implies \left(\frac{dy}{dx}ight)_1 = \frac{2a}{y} \dots (i)$For the curve $y = e^{-x/2a}$,differentiating with respect to $x$ gives:$\left(\frac{dy}{dx}ight)_2 = e^{-x/2a} \left(-\frac{1}{2a}ight) = -\frac{y}{2a} \dots (ii)$Two curves intersect orthogonally if the product of their slopes at the point of intersection is $-1$:$\left(\frac{dy}{dx}ight)_1 	imes \left(\frac{dy}{dx}ight)_2 = \left(\frac{2a}{y}ight) 	imes \left(-\frac{y}{2a}ight) = -1$Since the product is $-1$,the curve $y = e^{-x/2a}$ cuts the parabola $y^2 = 4ax$ at right angles.

Which one of the following curves cuts the parabola $y^2 = 4ax$ at right angles?

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