Write-up 11

by

Lynne Bombard

Problem: Investigate with different values of p,

, ,,,

for n>1, n=1, and n<1.


Step I

For the graph of , k=1 the graph is a parabola that opens to the right and does not matter what the value of p is. When the absolute value of p increases the graph is wider. The vertex of this parabola is (-p/2,0). And as we change the value of p, the y-intercepts are p and -p. When p is negative, the parabola is reflected over the y-axis.

When |n| >1, the graph is is a hyperbola. A negative value of p reflects the graph over the y-axis. To obtain the vertices of the hyperbola, we must evaluate when theta is pi and 2pi.

When -1<n<1, the graph is an ellipse.


For the graph , its just a reflection across the y-axis of everything that occurred in step I for the equation .


For the graph of , k=1 the graph is a parabola that opens upwards and it does not matter what the value of p is. When the absolute value of p increases the graph is wider. The vertex of this parabola is (0,-p/2). And as we change the value of p, the x-intercepts are p and -p. When p is negative, the parabola is reflected over the x-axis.

When |n| >1, the graph is is a hyperbola. A negative value of p reflects the graph over the x-axis. To obtain the vertices of the hyperbola, we must evaluate when theta is pi/2 and 3pi/2.

When -1<n<1, the graph is an ellipse.


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