The equation of any normal to the parabola {y | vedclass

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(D) The given parabola is ${y^2} = 4a(x - a)$.Let the point on the parabola be $(x_1, y_1)$. The equation of the tangent at $(x_1, y_1)$ is $y{y_1} =

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$y = m(x - a) - 2am - a{m^3}$

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(D) The given parabola is ${y^2} = 4a(x - a)$.Let the point on the parabola be $(x_1, y_1)$. The equation of the tangent at $(x_1, y_1)$ is $y{y_1} = 2a(x + x_1 - 2a)$.The slope of the tangent is $m_t = \frac{2a}{y_1}$.The slope of the normal is $m = -\frac{1}{m_t} = -\frac{y_1}{2a}$,which implies $y_1 = -2am$.Since the point $(x_1, y_1)$ lies on the parabola,${y_1^2} = 4a(x_1 - a)$.Substituting $y_1 = -2am$,we get $(-2am)^2 = 4a(x_1 - a)$,which simplifies to $4a^2m^2 = 4a(x_1 - a)$,so $x_1 - a = am^2$,or $x_1 = a + am^2$.The equation of the normal at $(x_1, y_1)$ is $y - y_1 = m(x - x_1)$.Substituting $x_1$ and $y_1$,we get $y - (-2am) = m(x - (a + am^2))$.$y + 2am = mx - am - am^3$.$y = m(x - a) - 2am - am^3$.

The equation of any normal to the parabola ${y^2} = 4a(x - a)$ is

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