The length of the subnormal to the parabola y | vedclass

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(D) Let the coordinates of point $P$ on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.The slope of the tangent at $P$ is given by differentiating $y^2 = 4

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$2a$

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(D) Let the coordinates of point $P$ on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.The slope of the tangent at $P$ is given by differentiating $y^2 = 4ax$ with respect to $x$:$2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$.The slope of the normal at $P$ is $-t$.The equation of the normal at $P(at^2, 2at)$ is:$y - 2at = -t(x - at^2)$$y - 2at = -tx + at^3$$tx + y = 2at + at^3$.To find the $x$-intercept $N$,set $y = 0$:$tx = 2at + at^3 \implies x = 2a + at^2$.Let $M$ be the projection of $P$ on the $x$-axis,so $M$ is $(at^2, 0)$.The length of the subnormal is the distance $MN = |x_N - x_M| = |(2a + at^2) - at^2| = 2a$.

The length of the subnormal to the parabola $y^2 = 4ax$ at any point is equal to

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