# Thread: Analytic Geometry: Ellipse [2]

1. ## Analytic Geometry: Ellipse [2]

Question:
Find the coordinates of the foci, the end points of the major axes and minor axis, the length of latus rectum and the end points of latus rectum.

$\displaystyle \frac{(x-3)^2}{16} + \frac{(y-2)^2}{9} = 1$

Attempt:

Center: $\displaystyle (3,2)$ $\displaystyle a=4$, $\displaystyle b=3$ , $\displaystyle c=\sqrt{16-9} = \sqrt{7}$

Foci:$\displaystyle (3-\sqrt{7},2)$ and $\displaystyle (3+\sqrt{7},2)$

End points of major axis: $\displaystyle (-1,2)$ and $\displaystyle (7,2)$

End points of minor axis: $\displaystyle (3,-1)$ and $\displaystyle (3,5)$

Length of latus rectum: $\displaystyle 2\frac{b^2}{a} = 2\frac{(3)^2}{4} = \frac{9}{2}$

How can I find out the end points of latus rectum? and are my answers right tell now?

2. ## Ellipse

Hello looi76
Originally Posted by looi76
Question:
Find the coordinates of the foci, the end points of the major axes and minor axis, the length of latus rectum and the end points of latus rectum.

$\displaystyle \frac{(x-3)^2}{16} + \frac{(y-2)^2}{9} = 1$

Attempt:

Center: $\displaystyle (3,2)$ $\displaystyle a=4$, $\displaystyle b=3$ , $\displaystyle c=\sqrt{16-9} = \sqrt{7}$

Foci:$\displaystyle (3-\sqrt{7},2)$ and $\displaystyle (3+\sqrt{7},2)$

End points of major axis: $\displaystyle (-1,2)$ and $\displaystyle (7,2)$

End points of minor axis: $\displaystyle (3,-1)$ and $\displaystyle (3,5)$

Length of latus rectum: $\displaystyle 2\frac{b^2}{a} = 2\frac{(3)^2}{4} = \frac{9}{2}$

How can I find out the end points of latus rectum? and are my answers right tell now?
You've got everything right so far. To get the end-points of the latus rectum is very easy. You know that the semi-latus-rectum has length $\displaystyle \frac94$, so just move up and down through this distance from each of the foci, and you get to each of the end points. So one of them will be $\displaystyle (3 + \sqrt7, 2 + \frac94)$. I'll leave the rest to you.