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:

  1. Identify the Point on the Parabola:

    • The end of the latus rectum for a parabola $y^2 = 4ax$ is at the focus, which is $(a, 0)$.
  2. 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 $(a, 2a)$.
  3. 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.
  4. 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 $-1$.
  5. 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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