# Maximum distance of normal from the center

• Dec 27th 2009, 10:30 PM
Maximum distance of normal from the center
Normal is drawn at a variable point 'P' of an ellipse $
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
$

Find the maximum distance of the normal from the centre of the ellipse
• Dec 28th 2009, 05:09 AM
TKHunny
1) The question makes no sense. If a = b, all the Normals, extended both directions, CONTAIN the center.

2) Okay, I guess it is sufficiently clear that a and b are not equal. I still don't like it.

3) Can you build the Normals?

You'll need a few tools.

1) Find the general derivative of all points on the ellipse.
2) Write the general equation of the tangent line of all points on the ellipse.
3) If you can do #2, you really should be able to write the general equation of all Normals.
4) Calculate the distance of a line from the Origin.

Where shall we start?

Note: The derivative is WAY easier using an implicit method, rather than first solving for y.
• Dec 28th 2009, 06:21 AM
HallsofIvy
Quote:

Normal is drawn at a variable point 'P' of an ellipse $
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
$

Find the maximum distance of the normal from the centre of the ellipse

That ellipse can also be written as $b^2x^2+ a^2y^2= a^2b^2$. Then $2b^2x+ 2a^2y\frac{dy}{dx}= 0$ so $\frac{dy}{dx}= -\frac{a^2y}{b^2x}$.

The normal to the ellipse at any point $(x_0,y_0)$ can be written $y= \frac{b^2x_0}{a^2y_0}(x- x_0)+ y_0$.

Any line normal to that line, passing through the origin, is $y= -\frac{a^2y_0}{b^2x_0}x$.

Since the perpendicular to a line, through a point, is the shortest distance to that line, find the point where those two lines intersect and find the distance from that point to the origin.
• Dec 28th 2009, 03:07 PM
TKHunny
Or, of course, use the standard distance formula (point to line) and just write it down. At (0,0), it simplifies quite a bit.

$\frac{|a^{2}y_{0}^{2}-b^{2}x_{0}^{2}|}{\sqrt{(a^{2}y_{0})^{2}+(b^{2}x_{0 })^{2}}}$