Can someone help me simplify this ellipse equation?

AI Thread Summary
To simplify the ellipse equation 15(x+2)^2 + (y+3)^2/4 = 4, divide both sides by 4 to achieve a standard form with 1 on the right side. This results in (15/4)(x+2)^2 + (1/4)(y+3)^2 = 1. To handle the coefficient of 15, recognize that it can be expressed as multiplying by 1/15, allowing for easier manipulation. The goal is to identify the values of a^2 and b^2, as well as the center coordinates (h, k). This approach will help in graphing the ellipse correctly.
iamsmooth
Messages
103
Reaction score
0

Homework Statement


Graph the following:
15(x+2)^2 + \frac{(y+3)^2}{4} = 4


Homework Equations


\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1


The Attempt at a Solution



I can't seem to get both coefficients to 1, since the right side is only 4. If I multiply the whole thing by 4, I can get rid of the denominator on the y term, but then the coefficient of the x jumps to 60, and 60 won't reduce the coefficient to 1. So I'm stuck. Can anyone help? This is not homework, just a practice question that's bugging me...
 
Physics news on Phys.org
You want to divide both sides by 4 so that you get a 1 on right hand side. Now you only need to collect the denominators into the form a2 and b2 and identify what h and k is. If you have trouble with the 15 then notice that multiplying something with 15 is the same as dividing by 1/15.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top