Finding the Line Normal to a Hyperbola: Domenic's Q&A at Yahoo Answers

[MarkFL]

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]

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:

View attachment 1658