Find the equation of the normal to the parabola y^2 = 4ax at the end of the latus rectum in the first quadrant.
Question
Find the equation of the normal to the parabola $y^2 = 4ax$ at the end of the latus rectum in the first quadrant.
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Answer
To find the equation of the normal to the parabola $y^2 = 4ax$ at the end of the latus rectum in the first quadrant, follow these steps:

Identify the Point on the Parabola:
 The end of the latus rectum for a parabola $y^2 = 4ax$ is at the focus, which is $(a, 0)$.

Determine the Coordinates:
 For the parabola $y^2 = 4ax$, its latus rectum's endpoints are $(a, \pm 2a)$. In the first quadrant, the relevant point is $(a, 2a)$.

Form the Equation of Normal:
 The standard form of the normal to a parabola $y^2 = 4ax$ at the point $(a, 2a)$ can be written as: $$ y = mx  2am  am^3 $$
 Here, $m$ is the slope of the normal.

Simplify and Solve:
 We need to simplify the equation at the point $(a, 2a)$. Recall that the slope $m$ for the normal at the latus rectum point has a unique value. Using known results, we can state that the slope $m$ is $1$.

Substitute the Slope:
 Substituting $m = 1$ into the equation: $$ y = 1 \cdot x  2a(1)  a (1)^3 $$
 This simplifies to: $$ y = x + 2a + a $$ $$ y = x + 3a $$
Thus, the equation of the normal to the parabola $y^2 = 4ax$ at the end of the latus rectum in the first quadrant is:
$$ \boxed{y = x + 3a} $$
This equation describes how the normal line behaves at the specified point on the given parabola.
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