Start by converting your equations to standard form. The first one is already in standard form. For the second one, you need to divide all terms by 36 so that the left-hand side equals one.
Then, identify your center and foci. For more information on that, there is a very good website. (Unfortunately, I'm unable to post links.) It's called "Pauls Online Notes: Algebra - Ellipses." Typing that into Google should bring it right up.
If you have any questions, just ask.
I need help with the following questions please.
Graph the ellipse.
((x-1)^2)/(9)+ ((y-2)^2)/(4)=1
(4(x+1)^2)+9(y-2)^2=36
3 answers
You already have the equation in the form
(x-x')^2/a^2 + (y-y')^2/b^2 = 1
That form tells you all you need to know about what the ellipse looks like
The center of the ellipse is at x=x'= 2. The major axis is along the y=y' ine and has a half-length of a=3 since the denominator under (x-1)^2 is 3^2). The minor axis is along the x=1 line and has a half-length of b=2 (from the 2^2 in the denominator under (y-2)^2.
You should also try computing a few points yourself. Assume a value of x-1 between -3 and +3 (x between -2 and 4), and compute the corresponding value(s) of y to get points on the ellipse.
(x-x')^2/a^2 + (y-y')^2/b^2 = 1
That form tells you all you need to know about what the ellipse looks like
The center of the ellipse is at x=x'= 2. The major axis is along the y=y' ine and has a half-length of a=3 since the denominator under (x-1)^2 is 3^2). The minor axis is along the x=1 line and has a half-length of b=2 (from the 2^2 in the denominator under (y-2)^2.
You should also try computing a few points yourself. Assume a value of x-1 between -3 and +3 (x between -2 and 4), and compute the corresponding value(s) of y to get points on the ellipse.
Correction: The center of the ellipse is at (x',y') = (1,2)
1 answer