Show that the line with equation y=2x+5 is a tangent to the ellipse with equation 9x^2+4y^2=36.

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- Mar 13th 2008, 07:44 PMrednestCoordinate 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 PMSoroban
Hello, rednest!

Quote:

Show that the line with equation $\displaystyle y\:=\:2x+5\;\;{\color{blue}[1]}$

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

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.