SOLUTION: The circle x^2 + y^2 - 2x - 3 = 0 is stretched horizontally by a factor of 2 about x = 0 to obtain an ellipse. what is the equation of this ellipse in general form

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The circle x^2 + y^2 - 2x - 3 = 0 is stretched horizontally by a factor of 2 about x = 0 to obtain an ellipse. what is the equation of this ellipse in general form      Log On


   

Question 437902: The circle x^2 + y^2 - 2x - 3 = 0 is stretched horizontally by a factor of 2 about x = 0 to obtain an ellipse. what is the equation of this ellipse in general form
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!


To stretch horizontally by a factor of k, the rule is:

Replace x by x/k

So we replace x by x/2 in

        xｲ + yｲ - 2x - 3 = 0

(x/2)ｲ + yｲ - 2(x/2) - 3 = 0

       xｲ/4 + yｲ - x - 3 = 0

Cl;earing of fractions:

      xｲ + 4yｲ - 4x - 12 = 0

That's all you were supposed to do.  However if you
place the circle and the ellipse in standard form
you get:

(x - 1)ｲ + (y - 0)ｲ = 2ｲ

which is this graph:



And the ellipse has the standard equation

(x - 2)ｲ   (y - 0)ｲ
覧覧覧覧 + 覧覧覧覧 = 1
   4ｲ         2ｲ

and its graph is:



Edwin