# Thread: Another Coordinate system!

1. ## Another Coordinate system!

Question

Prove that the line with equation $\displaystyle y = mx + c$ is a tangent to the ellipse with equation $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ if, and only if, $\displaystyle a^2m^2 + b^2 = c^2$.

HELP!

2. The line $\displaystyle y = m\cdot x + C$ is tangent to $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

There are two conditions:
1. The line and the ellipse must have a common point (x,y)
2. Their slopes must be equal at (x,y)

The slope of $\displaystyle y = m\cdot x + C$ is m.

And the ellipses is,
$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
(Implicit differentiation)
$\displaystyle \frac{2x}{a^2} + \frac{2yy'}{b^2} = 0$
(Some algebra)
$\displaystyle y'= -\frac{x b^2}{y a^2}$

As the slopes are equal, we can write
$\displaystyle y'=m= -\frac{x b^2}{y a^2}$

$\displaystyle y= -\frac{x b^2}{m a^2}$

Plug this is $\displaystyle y=m\cdot x + C$

$\displaystyle -\frac{x b^2}{m a^2}=m\cdot x + C$

$\displaystyle x= -\frac{m c a^2}{m^2 a^2 + b^2}$

Then plug it in $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\displaystyle \frac{x^2}{a^2} + \frac{x^2 b^2}{m^2 a^4} = 1$

$\displaystyle x^2 = \frac{1}{\frac{1}{a^2}+\frac{b^2}{m^2 a^4}}$

$\displaystyle x^2 = \frac{m^2 a^4}{m^2 a^2 + b^2}$

Plug the x we found before,

$\displaystyle {\left ( -\frac{m c a^2}{m^2 a^2 + b^2} \right )}^2 = \frac{m^2 a^4}{m^2 a^2 + b^2}$

$\displaystyle \frac{\not{m^2} c^2 \not{a^4}}{(m^2 a^2 + b^2)^{\not 2}} = \frac{\not{m^2} \not{a^4}}{\not m^2 \not a^2 + \not b^2}$

$\displaystyle \boxed{~c^2 = a^2 m^2 + b^2~}$