# Coordinate systems help!

• Mar 13th 2008, 07:44 PM
rednest
Coordinate systems help!
Show that the line with equation y=2x+5 is a tangent to the ellipse with equation 9x^2+4y^2=36.
• Mar 13th 2008, 08:56 PM
Soroban
Hello, rednest!

Quote:

Show that the line with equation $\displaystyle y\:=\:2x+5\;\;{\color{blue}[1]}$
is a tangent to the ellipse with equation $\displaystyle 9x^2+4y^2\:=\:36\;\;{\color{blue}[2]}$

First, find their intersection(s).

Substitute [1] into [2]: .$\displaystyle 9x^2 + 4(2x+5)^2 \:=\:36$

. . which simplifies: .$\displaystyle 25x^2 + 80x + 64 \:=\:0$

. . and factors: .$\displaystyle (5x + 8)^2\:=\:0$

. . and has one root: .$\displaystyle x \:=\:-\frac{8}{5}$
Substitute into [1] and we get: .$\displaystyle y \:=\:\frac{9}{5}$
The line and ellipse have one common point: .$\displaystyle \left(-\frac{8}{5},\:\frac{9}{5}\right)$

That should be sufficient, but let's check their slopes.

Obviously, the line has slope 2.

Differentiate [2] implicitly: .$\displaystyle 18x + 8y\,\frac{dy}{dx} \:=\:0\quad\Rightarrow\quad \frac{dy}{dx} \:=\:-\frac{9x}{4y}$

At $\displaystyle \left(-\frac{8}{5},\:\frac{9}{5}\right)$, the tangent to the ellipse has slope: .$\displaystyle \frac{dy}{dx} \;=\;-\frac{9\left(-\frac{8}{5}\right)}{4\left(\frac{9}{5}\right)} \;=\;-\frac{-\frac{72}{5}}{\frac{36}{5}} \;=\;2$

. . Their slopes are equal!

Therefore, the line is tangent to the ellipse.