The equation of the normal to the parabola y^ | vedclass

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Question

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

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$m^3y + m^2x = 2am^2 + a$

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(C) The equation of the parabola is $y^2 = 4ax$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,so $\frac{dy}{dx} = \frac{2a}{y}$.At the point $P\left( \frac{a}{m^2}, \frac{2a}{m} \right)$,the slope of the tangent is $m_t = \frac{2a}{2a/m} = m$.The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{m}$.The equation of the normal is $y - y_1 = m_n(x - x_1)$.$y - \frac{2a}{m} = -\frac{1}{m} \left( x - \frac{a}{m^2} \right)$.Multiplying by $m^3$,we get $m^3y - 2am^2 = -m^2x + a$.Rearranging the terms,we get $m^3y + m^2x = 2am^2 + a$.

The equation of the normal to the parabola $y^2 = 4ax$ at the point $\left( \frac{a}{m^2}, \frac{2a}{m} \right)$ is:

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