Another Coordinate system!

**rednest** · March 25th 2008, 05:55 AM

Question

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

HELP!

**wingless** · March 25th 2008, 08:12 AM

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

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 $y = m\cdot x + C$ is m.

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

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

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

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

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

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

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

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

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

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

Plug the x we found before,

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

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

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

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