Thread: Parametric equation of an ellipse

Can somebody please help me if the following is correct:

I must give parametric equations for the following ellipses:

1) 3x^2 + 5y^2 = 75

Since a=5 and b=sqrt(15), is this my correct answer:
x = 5 cos t
y = sqrt(15) sin t

2) 3x^2 + 12x + 5y^2 -10y -58 =0

Since a=5 and b=sqrt(15) like above mentioned, but this ellipse is transformed so that the new center became (-2, 1), then is it that the parametric equations became:
x= -2 + (5 cos t)
y= 1 + (sqrt(15) sin t)

Many thanks to someone who helps!
Brigitte

I got the same answers, so your answers look correct to me

Originally Posted by dolkam

Can somebody please help me if the following is correct:

I must give parametric equations for the following ellipses:

1) 3x^2 + 5y^2 = 75

Since a=5 and b=sqrt(15), is this my correct answer:
x = 5 cos t
y = sqrt(15) sin t

2) 3x^2 + 12x + 5y^2 -10y -58 =0

Since a=5 and b=sqrt(15) like above mentioned, but this ellipse is transformed so that the new center became (-2, 1), then is it that the parametric equations became:
x= -2 + (5 cos t)
y= 1 + (sqrt(15) sin t)

Many thanks to someone who helps!
Brigitte

You should be checking your answer by substituting the parametric equation into the cartesian ones. They are both correct.

$\displaystyle 3x^2 + 5y^2 = 75 $
Subbing in $\displaystyle x = 5 \cos t$ and $\displaystyle y = \sqrt{15} \sin t$
$\displaystyle = 75 \cos^2 t + 75 sin^2 t $
$\displaystyle = 75 (cos^2 t + sin^2 t) = 75 $


$\displaystyle 3x^2 + 12x + 5y^2 -10y -58 =0 $
$\displaystyle 3(x + 2)^2 + 5(y-1)^2 = 58 + 12 + 5 $
Subbing in $\displaystyle x = -2 + 5 \cos t$ and $\displaystyle y = 1 + \sqrt{15} \sin t$
$\displaystyle = 75 \cos^2 t + 75 sin^2 t $
$\displaystyle = 75 (cos^2 t + sin^2 t) = 75 $

Thank you very much.

I did not think about the substituting into the cartesian.

I will do that.

Brigitte