Question

How do I determine the shape of a graph by looking at the equation? (circle, ellipse, parabola, hyperbola) I know that if it's X^2-Y^2, it's a hyperbola, but how do you tell a circle apart from an ellipse or a parabola? Thank you in advance :). There are some really helpful people here. Much appreciated!

Answer 1

maybe if you could also show some of the rules for each... that would really help :p

Answer 2

This is an equation for an ellipse:
\[(x^2)/a^2 + (y^2)/b^2 = 1\]
This is the equation for a circle with radius r:
\[x^2 + y^2 = r^2\]

really, you can think of a circle as a special condition of an ellipse (just as a square is a special condition of a rectangle). that is, a circle can be thought of as an ellipse with equal radius all around. if this were the case, a and b in the ellipse equation would be equal to eachother. lets say they are equal to eachother and call them r. now we have:
\[(x^2)/r^2 + (y^2)/r^2 = 1\]
multiply both sides by r^2 and we get the equation for a circle!:
\[x^2 + y^2 = r^2\]

if we want to stretch the sides of the ellipse so that they are uneven, we can put in different values for a and b. for example, an ellipse with radius 3 in the x direction and radius 6 in the y direction would have the equation:
\[(x^2)/3^2 + (y^2)/6^2 = 1\]



also, all of these equations assume that you want your circle or ellipse to be centered at the origin (0,0) on your graph. you can offset your shape by adding or subtracting from the x or y. For example, if we wanted to make a circle that was centered at the point (2,0) we would just subtract 2 from x:
\[(x-2)^2 + y^2 = r^2\]

Answer 3

you might also see an ellipse written like so:
\[4x^2 + 9y^2 = 36\]

remember that in the normal equation for an ellipse we need to have a 1 to the right of the equals sign so to get a 1 there, we divide both sides by 36 and we get:
\[4x^2/36 + 9y^2/36 = 1\]\[x^2/9 + y^2/4 = 1\]\[x^2/3^2 + y^2/2^2 = 1\]

so this is really just an ellipse with radius 3 in the x direction and radius 2 in the y direction.

Answer 4

thank you so much! Mashimaro, you are awesome. You have a new fan :p