Asked by kevin
How would you draw the graph of this x^2/16 + y^2 = 1 ( this is an ellipse if I'm not mistaken)
I'm confused since y^2 has no value beneath it.
so what would the coordinates be for this.
the same problem with this one x^2/16 - y^2 = 1 ( a hyperbola correct)
if you can give me the coordinates for the graph as I have to plot them urgently and i'm a bit confused with the y^2 bit.
You can always draw a graph point by point by making up a table of x and y values. You are correct that the first equation is an ellipse. There will only be real values of y when x is between 4 and -4. The ellipse will be centered on the origin (0,0) and the y value will be betwen -1 and 1. When there is no denominator for the y^2 term in the general equation,
x^2/a^2 + y^2/b^2 = 1,
you can assume it to be b^2 = 1
The second equation is an hyperbola. There will only be real y values when x<-4 or x>4. I suggest you do the plotting. There will probably be asymptotes of x = 4y and x = -4y for large |x|, since the 1 then becomes negligible.
I'm confused since y^2 has no value beneath it.
so what would the coordinates be for this.
the same problem with this one x^2/16 - y^2 = 1 ( a hyperbola correct)
if you can give me the coordinates for the graph as I have to plot them urgently and i'm a bit confused with the y^2 bit.
You can always draw a graph point by point by making up a table of x and y values. You are correct that the first equation is an ellipse. There will only be real values of y when x is between 4 and -4. The ellipse will be centered on the origin (0,0) and the y value will be betwen -1 and 1. When there is no denominator for the y^2 term in the general equation,
x^2/a^2 + y^2/b^2 = 1,
you can assume it to be b^2 = 1
The second equation is an hyperbola. There will only be real y values when x<-4 or x>4. I suggest you do the plotting. There will probably be asymptotes of x = 4y and x = -4y for large |x|, since the 1 then becomes negligible.
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