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( *

Thanks in advance for leaving a reply :)

-Mj

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Re: Coordinate Geometry Problems

Quote:

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( *

Thanks in advance for leaving a reply :)

-Mj

Good morning,

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

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

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

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

3. The distance of a focus to the center is

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

4. According to the question you have to solve the system of equations:

$\displaystyle \left| \begin{array}{r}\frac{2^2}{a^2} + \frac{3^2}{b^2}=1 \\ \\ 3 \cdot \sqrt{a^2-b^2} = \frac{2b^2}a \end{array} \right.$

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

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

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

This looks a little bit terrifying. I'll show you the first steps.

Under the assumption that a must be positive, you can do:

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

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

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

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

1 Attachment(s)

Re: Coordinate Geometry Problems

hello,

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

Attachment 30518

Thanks for the reply sir !!

Re: Coordinate Geometry Problems

Quote:

Originally Posted by

**mventurina1** hello,

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

Attachment 30518
Thanks for the reply sir !!

Hello,

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

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

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

Multiply through the complete equation by a:

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

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

Divide the complete equation by 3aČ:

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

For the following steps see my previous posts.

Re: Coordinate Geometry Problems

Quote:

Originally Posted by

**earboth** Hello,

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

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

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

Multiply through the complete equation by a:

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

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

Divide the complete equation by 3aČ:

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

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))