# Semi Latus Rectum Proof - b^2/a

1. Feb 11, 2005

### sportsguy3675

I need to prove that the length of the semi latus rectum = $$b^2/a$$

I am assuming that you make some sort of a triangle within the ellipse to prove this, can someone point out which one it is?

2. Feb 12, 2005

### sportsguy3675

Why has no one responded? Doesn't anyone know how to do this?

3. Feb 12, 2005

### Andrew Mason

The equation for an ellipse relation is:

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

If you draw the ellipse with a string fixed at each foci, the length of the string is the major axis, 2a. The minor axis (x=0) is 2b. The ratio of the distance between the two foci to the length of the major axis is e, the eccentricity of the ellipse. So the distance between the two foci is 2ae which means that the foci are located at x=ae and x=-ae.

Now the semi-latus rectum is the line perpendicular to the major axis through one of the foci, to the ellipse. This is just the line: x=ae or x=-ae. The intersection of the line x=ae with the equation for the ellipse defines the coordinates of the end point of the semi-latus rectum. The y coordinate is, therefore its length.

So plug in the value x=ae for the ellipse relation and solve for y. That will give you l in terms of b. Then see if you can work it out from the relationships between a, b and e.

AM

4. Feb 13, 2005

### sportsguy3675

So I just do:
$$\frac{(ae)^2}{a^2} + \frac{y^2}{b^2} = 1$$

But what I don't get is what happens to y if you said to solve for y, but the answer is in terms of b?

5. Feb 13, 2005

### Andrew Mason

y is l.

(1)$$y^2 = l^2 = b^2(1-e^2)$$.

Use the Pythgorean relationship between b, the semi-major axis, and ae (the distance from the origin to a focus): $(ae)^2+b^2 = a^2$ (recall that a is 1/2 the length of the string which in this triangle is the hypotenuse with sides b and ae). Thus $b^2 = a^2(1-e^2)$ so $1-e^2 = b^2/a^2$.

Substituting $1-e^2 = b^2/a^2$ into (1) gives you:

$$l = \frac{b^2}{a}$$

AM

6. Feb 13, 2005

### sportsguy3675

Ok, I'll just write out my whole proof here because I like the equations better than my handwriting.

$$e = \frac{c}{a}$$

so $$c = ae$$

The line of the Semi Latus Rectum is $$x = \pm ae$$.
So you plug this in for x in the equation:

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

$$\frac{(ae)^2}{a^2} + \frac{y^2}{b^2} = 1$$

$$\frac{a^2e^2}{a^2} + \frac{y^2}{b^2} = 1$$

$$e^2 + \frac{y^2}{b^2} = 1$$

$$\frac{y^2}{b^2} = 1 - e^2$$

(1) $$y^2 = l^2 = b^2(1-e^2)$$

Plug $$c = ae$$ into $$a^2 - b^2 = c^2$$

$$c^2 + b^2 = a^2$$

$$(ae)^2 + b^2 = a^2$$

$$b^2 = a^2 - (ae)^2$$

$$b^2 = a^2(1 - e^2)$$

$$1 - e^2 = \frac{b^2}{a^2}$$

Plug back into (1).

$$l^2 = b^2(\frac{b^2}{a^2})$$

$$l^2 = (\frac{b^4}{a^2})$$

$$l = \frac{b^2}{a}$$

Thanks for the help Andrew!

Last edited: Feb 13, 2005