X^2/a^2 + Y^2/b^2 = 1
a = 2
b = 4
X^2/4 + Y^2/16 = 1
multiply through by 4
X^2 + Y^2/4 = 4
Y^2/4 = 4 - X^2
Y^2 = 4(4 - X^2)
Y^2 = 16 - 4X^2
Y = +/- sqrt(16 - 4X^2)
one of many ellipses possible ( don't for get to enter bot + and - terms )
any graph that is not represented by a line,ie: parabola, hyperbola, circle, ellipse,etc
No. It can also be a circle, ellipse or hyperbola.
As you may or may not be aware, there are multiple co-ordinate systems by which a graph may be defined. An ellipse graph has the general equation in the following systems:Cartesian (what most people are used to): (X-H)^2 (Y-K)^2---------- + ----------- = 1 (A,B,H,K are constants) A^2 B^2Polar: r(θ)= sqrt( (bcos(θ))^2+(asin(θ))^2 )Parametric: x = a cos(t) , y = b sin(t)
It is a straight line with gradient -A/B and intercept C/B.
ax^2+by^2=k is an ellipse this is not in standard form which is x^2/a^2+y^2/b^2=1 but you will often see ellipses written this way. ellipses are also commonly written in their parametric form which is x=ccos(t) and y=dsin(t). finally a circle is a special case of an ellipse and if a b or k are 0 or negative it is not an ellipse. c and d can be positive but not 0.
On my graphing calculator, a TI84 Plus, I can enter the equation into the Y= (a button) and then graph it by hitting the Graph button.
The whole ellipse shifts down by 6 units.
If a = b then it is a circle; otherwise it is an ellipse.
No, the graph of an oval/ellipse is not a function because it does not pass the vertical line test.
any graph that is not represented by a line,ie: parabola, hyperbola, circle, ellipse,etc
No. It can also be a circle, ellipse or hyperbola.
Yes. It's the graph of [ Y = f(X) ] described by (X/A)2 + (Y/B)2 = C2 A, B, and C are constants. If 'A' and 'B' are both '1', then the graph is a circle with radius 'C'.
Since it is not a function, you need to solve of y and you will have TWO equations. A plus and a minus one. FOR EXAMPLE the ellipse x^2/4+y+2/16=1 you could solve it as y^2/16=1-x^2/4 then multiply by 16 y^2=16(1-x^2/4) Now you have y=+ Square root of (16(1-x^2/4)) AND y=-Square root( 16(1-x^2/4)) IF you enter both of those and graph them on the same graph you will have an ellipse.
You move the graph upwards by 2 units.
I believe that that isn't possible unless you have it ported to ti84 or you know how to. the file format is different, and the game is built for a separate calc.
You can do the equation Y 2x plus 3 on a graph. On this graph the Y would equal 5 and X would equal to 0.
y = -0.5x plus or minus any number