Hello Domenic,
We are given the hyperbola:
$$\frac{x^2}{4}-\frac{y^2}{9}=1$$
Let's multiply through by $36$ to obtain:
(1) $$9x^2-4y^2=36$$
Letting $x=4$, we find:
$$9(4)^2-4y^2=36$$
Divide through by 4:
$$36-y^2=9$$
$$y^2=27$$
Since we are told $y>0$, taking the positive root, we find:
$$y=3\sqrt{3}$$
Hence, the coordinates of the point are:
$$\left(4,3\sqrt{3} \right)$$
Now, to find the normal slope, we need to implicitly differentiate (1) with respect to $-y$ to get:
$$18x\left(-\frac{dx}{dy} \right)-8y(-1)=0$$
$$-\frac{dx}{dy}=-\frac{4y}{9x}$$
Thus, the slope of the normal line at the given point is:
$$\left. -\frac{dx}{dy} \right|_{(x,y)=\left(4,3\sqrt{3} \right)}=-\frac{4\left(3\sqrt{3} \right)}{9(4)}=-\frac{1}{\sqrt{3}}$$
Now, we have a point on the normal line and the slope, thus the point-slope formula yields:
$$y-3\sqrt{3}=-\frac{1}{\sqrt{3}}(x-4)$$
Multiply through by $-\sqrt{3}$:
$$-\sqrt{3}y+9=x-4$$
Arrange in the required standard form:
$$x+\sqrt{3}y-13=0$$
Here is a plot of the hyperbola and the normal line at the given point:
View attachment 1658
