determine all (x,y) of the line Normal to an Ellipse

Question

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 $$(3.745,2.106)$$

Answer 1

Obs: I'm assuming you want to know which points int the ellipse have a normal line with slope $-4$

You seem to be confusing the normal line with the tangent line. You have the equation $$ \frac{x^2}{16}+\frac{y^2}{36}=1\Rightarrow y=\pm \frac{3}{2}\sqrt{16-x^2}$$ So taking the derivative, we have two situations: $$y'=\mp \frac{3x}{2\sqrt{16-x^2}}$$ Now, here is where you got confused. The derivative give us the slope of the tangent line at the given point. So, as the tangent line is orthogonal to the normal line, we have: $$ m_n \cdot y' = -1 \Rightarrow m_n = \pm \frac{2\sqrt{16-x^2}}{3x}$$ Now, all you have to do is solve $m_n =-4$. That yields two equations:$$2\sqrt{16-x^2}\pm3x=0$$. The rest is up to you. You only have to input the result in the original function $y(x)$.

Answer 2

If you are looking for the two points of the ellipse where the slope of the normal lines equal $-4$ as well as the equations of the two lines tangent to the ellipse at those two points, then proceed as follows:

As you correctly deduce, $$\dfrac{dy}{dx}=-\dfrac{9x}{4y}$$

Therefore $$-\dfrac{9x}{4y}=\dfrac{1}{4}$$

at the points in question, so they lie on the line $$y=-9x$$

Substituting into the equation of the ellipse yields the two points

$$ \left(\pm\dfrac{4}{\sqrt{37}},\mp\dfrac{36}{\sqrt{37}}\right)$$

Using the point slope equation of a line, one obtains the equations tangent to the ellipse at those two points:

$$ y=\dfrac{1}{4}x\mp\sqrt{37}$$

respectively.