The slope of the normal at the point (at^2, 2 | vedclass

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(C) The equation of the parabola is $y^2 = 4ax$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \fr

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$-t$

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(C) The equation of the parabola is $y^2 = 4ax$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \frac{2a}{y}$. At the point $(at^2, 2at)$,the slope of the tangent is $m_T = \frac{2a}{2at} = \frac{1}{t}$. The slope of the normal $m_N$ is given by $m_N = -\frac{1}{m_T}$. Therefore,$m_N = -\frac{1}{1/t} = -t$.

The slope of the normal at the point $(at^2, 2at)$ of the parabola $y^2 = 4ax$ is

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