We plot the corresponding points on the graph below, indicating the t values corresponding to each point.

• You should determine the points corresponding to t = 2 and t = -2.

The DERIVE expression [ 2 t^3 + 1, t^2 - 1 ] stands for the two parametric equations.   When plotted, this expression gives a graph like that shown below.

Video Clip #01

We can express the equation of this curve in terms of just x and y by eliminating the parameter t.

• We solve the first equation for t, as indicated.
• We then substitute the result into the second equation to obtain y = ( (x - 1) / 2 ) ^ (2/3) -1.
• If this equation is plotted using DERIVE, it gives the same result as the plot of the parametric equations.

Video Clip #02

We can eliminate the parameter `theta from the equations in the figure below.

We know in advance from previous work that these equations defined an ellipse with semi-axes 6 in the x direction and 2 in the y direction.

We could make a table and plot points; if you have not done so in a similar exercise, it is recommended that you do so now (use `theta values 0, `pi/6, `pi/3 and `pi/2 to start with; from the resulting points the rest of the points are fairly obvious).

To eliminate `theta, we begin by solving the first equation for `theta.

We obtain `theta = cos^-1 ( x / 6 ), which we substitute into the second equation.

To evaluate the resulting sin ( cos^-1( x/6) ), we proceed as in the following figure (you may consult this figure and return, or you may accept the result quoted here and then see where it came from after following the rest of this solution).

Our result is that sin (cos^-1(x / 6) ) = `sqrt( 6^2 - x^2 ) / 6.

The expression cos^-1 ( x / 6 ) represents the angle whose cosine is x / 6.

• This angle could be the indicated angle of a triangle with adjacent side x and hypotenuse 6.
• The sine of this angle will therefore be the opposite side over the hypotenuse.
• The opposite side is easily found to be `sqrt ( 6^2 - x^2).
• Thus we have sin (cos^-1 ( x / 6 ) ) = `sqrt ( 6^2 - x^2 ) / 6.

Placing the x and y terms on the left-hand the constant terms on the right-hand side, we obtain the second line.

• Dividing both sides by the constant right-hand side we obtain the third line.
• Expressing the denominators as squares we obtain the standard form in the fourth line.

We see that the ellipse does indeed have semi-axes 6 and 2 in the x and y directions respectively.

Video Clip #03

In response to another question about completing the square, we determine the standard form of the equation indicated below.

• In the second line we prepare to complete to square by placing all constant terms of the right-hand side.
• In the third line we factor out the coefficients of the squared terms.
• In the fourth line we complete the squares within the parentheses.
• In the fifth line we factor our perfect squares.
• In the six line we distribute the factors.
• In the last line we move the resulting constants to the right-hand side.

In the figure below we first divide by the right-hand side to obtain 1 on that side.

In the second line we express the entire coefficient of each square as a denominator with a square.