椭圆二阶导数求解及法线与椭圆交点问题：二阶导数化简遇阻

Hey there! Let's work through your two ellipse problems step by step—first tackling that stuck second derivative, then moving on to finding normal line intersections.

You already nailed the first derivative: dy/dx = -x/(4y). Let's pick up from the intermediate step you had:

$$
y'' = \frac{-4y + \frac{-4x^2}{4y}}{16y2}
$$

First, clean up the numerator by combining terms over a common denominator. Multiply $-4y$ by $4y/4y$ to match the fraction's denominator:

$$
\text{Numerator} = \frac{-4y \cdot 4y}{4y} + \frac{-4x^2}{4y} = \frac{-16y^2 - 4x^2}{4y}
$$

Substitute this back into the second derivative expression:

$$
y'' = \frac{\frac{-16y^2 - 4x^2}{4y}}{16y2} = \frac{-16y^2 - 4x^2}{4y \cdot 16y^2}
$$

Factor out a $-4$ from the numerator and simplify the denominator:

$$
y'' = \frac{-4(4y^2 + x^2)}{64y3}
$$

Here's the key simplification trick: use the original ellipse equation $x^2 + 4y^2 = 4$ to replace the $4y^2 + x^2$ term in the numerator with 4:

$$
y'' = \frac{-4 \cdot 4}{64y^3} = \frac{-16}{64y^3} = -\frac{1}{4y^3}
$$

That's the fully simplified second derivative!

Let's start with a general point $(x_0, y_0)$ on the ellipse (so it satisfies $x_0^2 + 4y_0^2 = 4$).

Step 1: Find the normal line's slope #

The tangent slope at $(x_0, y_0)$ is your first derivative evaluated at that point: $m_{\text{tangent}} = -x_0/(4y_0)$. The normal line is perpendicular to the tangent, so its slope is the negative reciprocal:
$$
m_{\text{normal}} = \frac{4y_0}{x_0} \quad (\text{for } x_0 \neq 0; \text{ if } x_0=0, \text{ the normal is the y-axis, intersecting the ellipse at } (0,1) \text{ and } (0,-1))
$$

Step 2: Write the normal line equation #

Using point-slope form:
$$
y - y_0 = \frac{4y_0}{x_0}(x - x_0)
$$
Simplify to slope-intercept form:
$$
y = \frac{4y_0}{x_0}x - 3y_0
$$

Step 3:联立 with the ellipse equation #

Substitute $y$ from the normal line into $x^2 + 4y^2 = 4$. Instead of solving the full quadratic from scratch, we can use Vieta's theorem (since we know $(x_0, y_0)$ is one intersection point) to find the second intersection:

• The x-coordinate of the second point:
  $$
  x_1 = x_0 \cdot \frac{9y_0^2 - 1}{1 + 15y_0^2}
  $$
• The corresponding y-coordinate:
  $$
  y_1 = -\frac{y_0(9y_0^2 + 7)}{1 + 15y_0^2}
  $$

So the two intersection points are the original point $(x_0, y_0)$ and the new point $\left( x_0 \cdot \frac{9y_0^2 - 1}{1 + 15y_0^2}, -\frac{y_0(9y_0^2 + 7)}{1 + 15y_0^2} \right)$.

内容的提问来源于stack exchange，提问作者Jwan622