# Domenic's question at Yahoo! Answers regarding finding the line normal to a hyperbola

#### MarkFL

Staff member
Here is the question:

An Hyperbola has the equation x^2/4 - y^2/9 = 1?

- Find the coordinates of the point on this curve with x=4, y>0
- Find the slope of the normal to the curve at this point, and hence find the equation of the normal. Give the equation in general form, ie. Ax+By+C=0
I have posted a link there to this thread so the OP can see my work.

#### MarkFL

Staff member
Hello Domenic,

We are given the hyperbola:

$$\displaystyle \frac{x^2}{4}-\frac{y^2}{9}=1$$

Let's multiply through by $36$ to obtain:

(1) $$\displaystyle 9x^2-4y^2=36$$

Letting $x=4$, we find:

$$\displaystyle 9(4)^2-4y^2=36$$

Divide through by 4:

$$\displaystyle 36-y^2=9$$

$$\displaystyle y^2=27$$

Since we are told $y>0$, taking the positive root, we find:

$$\displaystyle y=3\sqrt{3}$$

Hence, the coordinates of the point are:

$$\displaystyle \left(4,3\sqrt{3} \right)$$

Now, to find the normal slope, we need to implicitly differentiate (1) with respect to $-y$ to get:

$$\displaystyle 18x\left(-\frac{dx}{dy} \right)-8y(-1)=0$$

$$\displaystyle -\frac{dx}{dy}=-\frac{4y}{9x}$$

Thus, the slope of the normal line at the given point is:

$$\displaystyle \left. -\frac{dx}{dy} \right|_{(x,y)=\left(4,3\sqrt{3} \right)}=-\frac{4\left(3\sqrt{3} \right)}{9(4)}=-\frac{1}{\sqrt{3}}$$

Now, we have a point on the normal line and the slope, thus the point-slope formula yields:

$$\displaystyle y-3\sqrt{3}=-\frac{1}{\sqrt{3}}(x-4)$$

Multiply through by $-\sqrt{3}$:

$$\displaystyle -\sqrt{3}y+9=x-4$$

Arrange in the required standard form:

$$\displaystyle x+\sqrt{3}y-13=0$$

Here is a plot of the hyperbola and the normal line at the given point: