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.
To draw the graph of the equation x^2/16 + y^2 = 1, which represents an ellipse, you can follow these steps:
1. Identify the center: The center of the ellipse is at the origin (0,0) since there are no additional terms in the equation.
2. Determine the major and minor axes: Since the equation is in the form x^2/a^2 + y^2/b^2 = 1, the major axis is along the x-axis and has a length of 2a, and the minor axis is along the y-axis and has a length of 2b. In this case, a = 4 and b = 1, so the major axis is 8 units long and the minor axis is 2 units long.
3. Plot the points: To plot the ellipse, start at the origin and move a distance of +/- a (4 units) along the x-axis and a distance of +/- b (1 unit) along the y-axis. For this equation, the points will be (4,0), (-4,0), (0,1), and (0,-1).
4. Draw the ellipse: Using the four plotted points, sketch a smooth curve that connects them, following the shape of an ellipse. Make sure the curve remains within the bounds of the axes.
For the second equation x^2/16 - y^2 = 1, which represents a hyperbola, follow these steps:
1. Identify the center: Similar to the ellipse, the center of the hyperbola is at the origin (0,0).
2. Determine the transverse and conjugate axes: For a hyperbola in the form x^2/a^2 - y^2/b^2 = 1, the transverse axis is along the x-axis and has a length of 2a, and the conjugate axis is along the y-axis and has a length of 2b. In this case, a = 4 and b = 1, so the transverse axis is 8 units long and the conjugate axis is 2 units long.
3. Draw the asymptotes: The asymptotes of a hyperbola are lines that intersect at the center of the hyperbola and have slopes of +/- (b/a). In this case, the asymptotes are x = 4y and x = -4y. Sketch these lines on the graph.
4. Plot additional points: Since the hyperbola is not bounded, you can plot more points along the curve to get a better understanding of its shape. Choose x-values that are larger than 4 or smaller than -4, and calculate the corresponding y-values using the equation y = +/- sqrt((x^2/16) - 1).
5. Draw the hyperbola: Using the asymptotes and the additional plotted points, sketch the curve of the hyperbola, making sure it follows the characteristic shape of a hyperbola.
These steps should help you accurately draw the graphs for both the ellipse and the hyperbola.