# Math Help - Coordinate Geometry Problems

1. ## Coordinate Geometry Problems

Hello!!,

Here's a sample problem: (Ellipse) (I have no idea)

"Passing through (2,3), Center at the origin, Latus rectum three times the distance from center to focus"

*i tried to manipulate the LR eqn (2b^2/a) and substituting the x&y to the eqn of the ellipse but I cant Get the formulas X( *

-Mj

2. ## Re: Coordinate Geometry Problems

Originally Posted by mventurina1
Hello!!,

Here's a sample problem: (Ellipse) (I have no idea)

"Passing through (2,3), Center at the origin, Latus rectum three times the distance from center to focus"

*i tried to manipulate the LR eqn (2b^2/a) and substituting the x&y to the eqn of the ellipse but I cant Get the formulas X( *

-Mj
Good morning,

1. The equation of an ellipse with center in the origin is

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

2. The coordinates of the given point satisfy the equation of the ellipse:

$\frac{2^2}{a^2} + \frac{3^2}{b^2}=1$

3. The distance of a focus to the center is

$f = \sqrt{a^2-b^2}$

4. According to the question you have to solve the system of equations:
$\left| \begin{array}{r}\frac{2^2}{a^2} + \frac{3^2}{b^2}=1 \\ \\ 3 \cdot \sqrt{a^2-b^2} = \frac{2b^2}a \end{array} \right.$

5. From the 1st equation you'll get: $b^2 = \frac{9a^2}{a^2-4}$

Sub in this term into the 2nd equation to replace b²:

$3 \cdot \sqrt{a^2-\frac{9a^2}{a^2-4}} = \frac{2 \cdot \frac{9a^2}{a^2-4}}a$

This looks a little bit terrifying. I'll show you the first steps.
Under the assumption that a must be positive, you can do:

$3 \cdot a^2 \cdot \sqrt{\frac{a^2-13}{a^2-4}} = {2 \cdot \frac{9a^2}{a^2-4}}$

$\sqrt{\frac{a^2-13}{a^2-4}} = \frac{6}{a^2-4}}$

$\sqrt{a^2-13} = \frac{6}{\sqrt{a^2-4}}}$

Now square both sides and solve for a. The only valid solution is $a = 4$, consequently $b = 2 \cdot \sqrt{3}$ and f = 2. That means the given point is the endpoint of the latus rectum.

3. ## Re: Coordinate Geometry Problems

hello,

i was just thinking about what happened to "a" as seen in this screen capture? %)

Thanks for the reply sir !!

4. ## Re: Coordinate Geometry Problems

Originally Posted by mventurina1
hello,

i was just thinking about what happened to "a" as seen in this screen capture? %)

Thanks for the reply sir !!
Hello,

sorry that I skipped an important step. Here we go (you remember: a must be positive!):

$3 \cdot \sqrt{a^2-\frac{9a^2}{a^2-4}} = \frac{2 \cdot \frac{9a^2}{a^2-4}}a$

$3 \cdot a \cdot \sqrt{1-\frac{9}{a^2-4}} = \frac{2 \cdot \frac{9a^2}{a^2-4}}a$

Multiply through the complete equation by a:

$3 \cdot a^2 \cdot \sqrt{1-\frac{9}{a^2-4}} = { \frac{18a^2}{a^2-4}}$

$3 \cdot a^2 \cdot \sqrt{\frac{a^2-4-9}{a^2-4}} = { \frac{18a^2}{a^2-4}}$

Divide the complete equation by 3a²:

$\sqrt{\frac{a^2-13}{a^2-4}} = \frac{6}{a^2-4}}$

For the following steps see my previous posts.

5. ## Re: Coordinate Geometry Problems

Originally Posted by earboth
Hello,

sorry that I skipped an important step. Here we go (you remember: a must be positive!):

$3 \cdot \sqrt{a^2-\frac{9a^2}{a^2-4}} = \frac{2 \cdot \frac{9a^2}{a^2-4}}a$

$3 \cdot a \cdot \sqrt{1-\frac{9}{a^2-4}} = \frac{2 \cdot \frac{9a^2}{a^2-4}}a$

Multiply through the complete equation by a:

$3 \cdot a^2 \cdot \sqrt{1-\frac{9}{a^2-4}} = { \frac{18a^2}{a^2-4}}$

$3 \cdot a^2 \cdot \sqrt{\frac{a^2-4-9}{a^2-4}} = { \frac{18a^2}{a^2-4}}$

Divide the complete equation by 3a²:

$\sqrt{\frac{a^2-13}{a^2-4}} = \frac{6}{a^2-4}}$

For the following steps see my previous posts.

Hi Sir, Thanks again for the reply now I clearly understand it.(wish I could pass the exams XX))