Hi everyone I have a question that requires me to determine the (x,y) coordinates of all points that intersects the x-axis on this ellipse when the normal line has a slope of -4, and I'm curious to know if I am doing this correctly, if so is my work correct and is there a way to check if it is?. Thank you all very much in advance.

Given Function:
$$ \frac {(x^2)}{16} + \frac{(y^2)} {36} = 1$$

I understand I take the derivative of the function implicitly or explicitly first, respectively they are:

$$y'=\frac{-9x} {{4y}}$$

$$y=\frac{-3x} {2\sqrt{16-x^2}}$$

First the implicit derivation and setting it equal to $-4$

$$-4 =\frac{-9x} {{4y}} $$

Then solving for $x$ and inputting the expression back into the original function

$$y = \frac{9x} {{16}} $$ 

$$\frac{x^2}{16} + {(\frac{9x}{16})^2\over{36}} = 1$$ 

$$\frac{x^2}{16} + \frac{81x^2}{256} \cdot \frac{1}{36} = 1$$

$$\frac{x^2}{16} + \frac{9x^2}{1024} = 1$$

$$\frac{73x^2}{1024}=1$$

$$x=\sqrt{\frac{1024}{73}}$$

$$ x=3.745 $$

Inputting the value of $x$ into the original function, solving for $y$

$$\frac{(3.745)^2}{16}+\frac{y^2}{36} = 1$$

$$y=\sqrt{1-0.8767\cdot{36}}$$

$$y=2.106$$

therefore the coordinates are $$(3.745,2.1067)$$

Now for the explicit derivation and setting it equal to $-4$

$$-4=\frac{-3x} {2\sqrt{16-x^2}}$$ 

solving for x

$$-4\cdot{2}=\frac{-3x}{\sqrt{16-x^2}}$$

$$(-8)^2 = \frac{9x^2}{16-2x^2}$$

$$(64)(16-2x^2)=9x^2$$

$$1024-64x^2 = 9x^2$$

$$-73x^2=-1024$$
$$x=\sqrt\frac{1024}{73}$$
$$x=3.745$$

Inputting the value of $x$ into the original function

$$\frac{x^2}{16}+\frac{y^2}{36} = 1$$
$$\frac{3.745^2}{16}+\frac{y^2}{36}=1$$ 
$$y=\sqrt{1-0.876}\cdot{36}$$
$$y=2.106$$

Therefore the coordinates are $$(3.745,2.106)$$