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Math Help - Coordinate Geometry Problems

  1. #1
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    Lightbulb 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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  2. #2
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    Re: Coordinate Geometry Problems

    Quote Originally Posted by mventurina1 View Post
    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

    \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 bČ:

    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.
    Attached Thumbnails Attached Thumbnails Coordinate Geometry Problems-ellipmitlatrectum.png  
    Thanks from mventurina1
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  3. #3
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    Re: Coordinate Geometry Problems

    hello,

    i was just thinking about what happened to "a" as seen in this screen capture? %)
    Coordinate Geometry Problems-ex.png

    Thanks for the reply sir !!
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  4. #4
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    Re: Coordinate Geometry Problems

    Quote Originally Posted by mventurina1 View Post
    hello,

    i was just thinking about what happened to "a" as seen in this screen capture? %)
    Click image for larger version. 

Name:	ex.png 
Views:	2 
Size:	19.7 KB 
ID:	30518

    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 3aČ:

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

    For the following steps see my previous posts.
    Thanks from mventurina1
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  5. #5
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    Re: Coordinate Geometry Problems

    Quote Originally Posted by earboth View Post
    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 3aČ:

     \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))
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