# A question.

• Dec 3rd 2006, 01:01 PM
Skor
A question.
I have this math problem that I am having trouble with, I really don't understand exactly what its asking for, or the best way to do it. I've tried a few different ways, but my answers never seemed right.

The question is.

"Find the equation of the line with a positive slope that is the tangent to the ellipse (x^2)/9 + (4^2)/4 = 1 at x=2.

??
• Dec 3rd 2006, 01:30 PM
topsquark
Quote:

Originally Posted by Skor
I have this math problem that I am having trouble with, I really don't understand exactly what its asking for, or the best way to do it. I've tried a few different ways, but my answers never seemed right.

The question is.

"Find the equation of the line with a positive slope that is the tangent to the ellipse (x^2)/9 + (y^2)/4 = 1 at x=2.

??

If it's got a positive slope at x = 2, then it's tangent to the bottom side of the ellipse, so let:
$y = - \sqrt{4 \cdot \left ( 1 - \frac{x^2}{9} \right ) }$ <-- This is the ellipse equation solved for y.

So
$y(2) = -\frac{2}{3} \sqrt{5}$
meaning the tangent line goes through the point $\left ( 2 , -\frac{2}{3} \sqrt{5} \right )$.

The tangent line will have a slope equal to the first derivative of y at x = 2:

$m = y' = -\frac{1}{2} \frac{1}{\sqrt{4 \cdot \left ( 1 - \frac{x^2}{9} \right )}} \cdot 4 \frac{-2x}{9} |_{x = 2}$

$m = \frac{2x}{9 \sqrt{1 - \frac{x^2}{9}}} |_{x = 2}$

$m = \frac{4}{3 \sqrt{5}} = \frac{4 \sqrt{5}}{15}$

So the tangent line is:
$y = \frac{4 \sqrt{5}}{15} x + b$

Using the point $\left ( 2 , -\frac{2}{3} \sqrt{5} \right )$ gives:
$-\frac{2}{3} \sqrt{5} = \frac{8 \sqrt{5}}{15} + b$

$b = -\frac{2}{3} \sqrt{5} - \frac{8 \sqrt{5}}{15} = - \frac{18 \sqrt{5}}{15}$

Thus the tangent line is:
$y = \frac{4 \sqrt{5}}{15} x - \frac{18 \sqrt{5}}{15}$

-Dan
• Dec 3rd 2006, 01:37 PM
Skor
Wow thanks for the detailed reply, I think I got this now, thanks alot.