Lesson Derive an equation and show that the locus of points is an ellipse

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Derive an equation and show that the locus of points is an ellipse


Problem 1

What is the equation of the locus of the point which moves so that its distance from the point  (2,0)
is  2/3  its distance from the line  y = 5 ?

Solution

Let (x,y) be the point of the locus.


Then the distance to point (2,0) is  d = sqrt%28%28x-2%29%5E2%2By%5E2%29,

while the distance to the line y=5 is  |y-5|.


Equate the equal distances


    sqrt%28%28x-2%29%5E2%2By%5E2%29 = %282%2F3%29%2Aabs%28y-5%29.   


It is the basic equation which translates the input into mathematical form.


Next, square both sides of this equation and reduce it to the standard conic section equation.


    %28x-2%29%5E2+%2B+y%5E2 = %284%2F9%29%2A%28y-5%29%5E2    

    x%5E2+-+4x+%2B+4+%2B+y%5E2 = %284%2F9%29%2A%28y%5E2+-+10y+%2B+25%29

    9x%5E2+-+36x+%2B+36+%2B+9y%5E2 = 4y%5E2+-+40y+%2B+100

    9x%5E2+-+36x +  5y%5E2+%2B+40y = 64

    9%2A%28x%5E2+-+4x+%2B+4%29 + 5%2A%28y%5E2+%2B+8y+%2B+16%29 = 64 + 36 + 80

    9%2A%28x-2%29%5E2 + 5%2A%28y%2B4%29%5E2 = 180.

    %28x-2%29%5E2%2F20 + %28y%2B4%29%5E2%2F36 = 1


It is the standard form equation of the ellipse centered at the point  (2,-4)  and

the major semi-axis  sqrt%2836%29 = 6 (vertical)  and the minor semi-axis  sqrt%2820%29 = 2%2Asqrt%285%29  (horizontal).


My other lessons in this site on deducing equations for locuses of points are

    - Derive an equation and show that the locus of points is a straight line
    - Derive an equation and show that the locus of points is a circle
    - Derive an equation and show that the locus of points is a hyperbola
    - Derive an equation and show that the locus of points is a parabola

    - OVERWIEW of lessons on deducing equations for locuses of points

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.


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