# Parametric equation of an ellipse

• April 10th 2010, 02:39 PM
dolkam
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
• April 10th 2010, 05:06 PM
Stroodle
I got the same answers, so your answers look correct to me :)
• April 10th 2010, 05:12 PM
Gusbob
Quote:

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.

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

$3x^2 + 12x + 5y^2 -10y -58 =0$
$3(x + 2)^2 + 5(y-1)^2 = 58 + 12 + 5$
Subbing in $x = -2 + 5 \cos t$ and $y = 1 + \sqrt{15} \sin t$
$= 75 \cos^2 t + 75 sin^2 t$
$= 75 (cos^2 t + sin^2 t) = 75$
• April 11th 2010, 12:05 AM
dolkam
Thank you very much.

I did not think about the substituting into the cartesian.

I will do that.

Brigitte
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