Show that the line with equation y=2x+5 is a tangent to the ellipse with equation 9x^2+4y^2=36.
Hello, rednest!
First, find their intersection(s).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.