Question 281498
Could you please show me how to do the following question because I got it wrong. The circle x^2 + y^2 - 2x - 3 =0 is stretched horizontally by a factor of 2 to obtain an ellispe. What is the equation of this ellipse in general form.Please help and thank you for your time.

putting equation into standard form of
{{{ (x-h)^2 + (y-k)^2 = r^2 }}}, where (h,k) is center of circle, and r is the radius
x^2 + y^2 - 2x - 3 = 0
x^2 - 2x + y^2 = 3
x^2 - 2x + 1 + y^2 = 4 (added 1 to both sides to complete the square)
{{{ (x - 1)^2 + y^2 = 4 }}}
or {{{ (x - 1)^2 + (y - 0)^2 = 4 }}}
center is (1,0) and radius is 2
another way to write this would be:
{{{ ((x - 1)^2)/4 + y^2/4 = 1 }}}
{{{ a = b = r = 2 }}}
now to stretch this cicle horizontally by a factor of 2 and write the equation of the ellipse
{{{ x^2/a^2 + y^2/b^2 = 1 }}}, {{{ a>b }}} is the equation we want for the ellipse
stretch horizontally by factor of 2 --> a=2 becomes a=4  --> now a^2=16
{{{ ((x - 1)^2)/16 + y^2/4 = 1 }}}
and that is the equation of the ellipse