MHB Equation of ellipse from directrix and focus

  • Thread starter Thread starter hatelove
  • Start date Start date
  • Tags Tags
    Ellipse Focus
AI Thread Summary
To find the equation of an ellipse with a directrix at y = -2 and a focus at the origin, the polar equation is derived using the relationship between the focus and directrix. The formula used is r = (pe)/(1 - sinθ), indicating the ellipse is elongated along the y-axis. The eccentricity 'e' is crucial for determining the ellipse's properties, but the center point and vertex need to be established first. The distances from a point on the ellipse to the focus and directrix are defined, leading to the condition for the ellipse: r/(r cos(θ) + 2) = e. The discussion emphasizes the need to find the major axis length and vertex positions to complete the equation.
hatelove
Messages
101
Reaction score
1
Find the equation of an ellipse with a directrix y = -2 and a focus at the origin.

I'm trying to find the polar equation first, and I learned this today but I forgot a lot of it and we're not allowed to take notes in class (professor says it helps to learn better) so I'm trying to look it up online but it's not much help because I can't find any elementary lessons on polar equations.

Anyway, I think the formula is

r = \frac{pe}{1 \pm sin\theta }

because of the position of the directrix, I know the ellipse is elongated along the y-axis. For this particular position given in the problem, I know it's a negative sign in the denominator (for the plus-minus symbol).

I know 'e' is eccentricity and to find that I need the center point but I forgot how to get it. I also need the vertex but I forgot how to find that too. But I know the lower vertex is between that given focus and directrix.

So far I have:

r = \frac{(2)e}{1 - sin\theta }

First step: how do I find the length of the major axes or a vertex?
 
Mathematics news on Phys.org
daigo said:
I'm trying to find the polar equation first, and I learned this today but I forgot a lot of it and we're not allowed to take notes in class (professor says it helps to learn better) so I'm trying to look it up online but it's not much help because I can't find any elementary lessons on polar equations.

Anyway, I think the formula is

r = \frac{pe}{1 \pm sin\theta }

because of the position of the directrix, I know the ellipse is elongated along the y-axis. For this particular position given in the problem, I know it's a negative sign in the denominator (for the plus-minus symbol).

I know 'e' is eccentricity and to find that I need the center point but I forgot how to get it. I also need the vertex but I forgot how to find that too. But I know the lower vertex is between that given focus and directrix.

So far I have:

r = \frac{(2)e}{1 - sin\theta }

First step: how do I find the length of the major axes or a vertex?

The directrix/forus definition of an ellipse is the locus of points such that the ratio of the distance from the focus to the distance from the directrx is a constant less than one.

Here the focus is the origin so the x-y co-ordinates of a general point on the ellipse is \( (r \cos(\theta), r \sin(\theta))\)m so the distance of a point on the ellipse from the focus is \(d_f=r\). The distance of the point from the directrix at \(y=-2\) is \(d_d=r\cos(\theta)+2\).

So the condition for this to be an ellipse is:

\[\frac{d_f}{d_d}=\frac{r}{r\cos(\theta)+2}=e\]

rearranging:

\[r=\frac{2e}{1-e\cos(\theta))}\]

The ends of the major axis correspond to \(\theta=0\) and \(\theta=\pi/2\) ,...

CB
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top