Two derivative problems. Stuck on finding the normal line.

Question

I have two problems. One is finding the second derivative. The other is finding an intersection of a normal line and an ellipsis.

1. Find $y''$ $$ x^2 + 4y^2 = 4$$

$$y' = 2x +4 \cdot 2y \cdot \frac{dy}{dx} = 0$$ $$2x + 8y \frac{dy}{dx} = 0$$ $$8y \frac{dy}{dx} = -2x$$ $$\frac{dy}{dx} = \frac{-2x}{8y} = \frac{-x}{4y}$$ $$ y'' = \frac{-4y - (-x \cdot 4 \frac{dy}{dx}}{16y^2}$$

$$ \frac{-4y + (4x\frac{dy}{dx})}{16y^2}$$

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

Now I'm stuck. I feel like I can simplify more?

2. Find an additional intersection point of the ellipsis with the normal line at (-1,1).

$$x^2 - xy + y^2 = 3$$ $$2x - x\frac{dy}{dx} + y + 2y \frac{dy}{dx} = 0$$ $$x \cdot \frac{dy}{dx} - 2y \cdot \frac{dy}{dx} = 2x + y$$ $$2x + y = \frac{dy}{dx} (x - 2y)$$ $$\frac{2x + y}{x - 2y} = \frac{dy}{dx}$$

So at (-1,1) the slope is $\frac{1}{3}$

So the slope of the normal is -3. But where do I go from here?

Answer 1

for the first derivative we get $$2x+8yy'=0$$ or $$y'=-\frac{x}{4y}$$ computing the derivative of the equation $$2x+8yy'=0$$ once more with respect to $x$ we get: and the second derivative $$2+8y'^2+8yy''=0$$ solve this for $y''$ and plug in the formula for $y'$ for #2) we get $$y=-3x+n$$ with $x=-1,y=1$ we can compute $n$ $n$ stands for the y-intercept of the linear function $$y=mx+n$$

Answer 2

You can simplify

$y'' = \frac{-4y + \frac{-4x^2}{4y}}{16y^2} = \frac{-4y^2 -x^2}{16y^3}$

In 2) you have a sign flipped in your calculation of $\frac {dy}{dx} = \frac {2x-y}{x-2y}$

at $(-1,1)$ the slope the tanget equals $1$ and the slope of the normal $= -1$

your line $(y-1) = -(x+1)$