Two parabolas y^2 = 4a(x - l_1) and x^2 = 4a( | vedclass

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(C) Let the point of contact be $(x_1, y_1)$.For the first parabola $y^2 = 4a(x - l_1)$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} =

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$xy = 4a^2$

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(C) Let the point of contact be $(x_1, y_1)$.For the first parabola $y^2 = 4a(x - l_1)$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4a$,so $\frac{dy}{dx} = \frac{2a}{y_1}$.For the second parabola $x^2 = 4a(y - l_2)$,differentiating with respect to $x$ gives $2x = 4a \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{x_1}{2a}$.Since the parabolas touch at $(x_1, y_1)$,their slopes must be equal at this point.Therefore,$\frac{2a}{y_1} = \frac{x_1}{2a}$.This implies $x_1 y_1 = 4a^2$.Thus,the locus of the point of contact is $xy = 4a^2$.

Two parabolas $y^2 = 4a(x - l_1)$ and $x^2 = 4a(y - l_2)$ always touch one another,where $l_1$ and $l_2$ are variable. The locus of their point of contact has the equation:

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