# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2003 | May-Jun | (P2-9709/02) | Q#7

Question

The parametric equations of a curve are

i. Show that

ii. Find the equation of the tangent to the curve at the point where .

iii. For the part of the curve where , find the coordinates of the points where the tangent  is parallel to the x-axis.

Solution

i.

We are required to show that  for the parametric equations given below;

If a curve is given parametrically by equations for  and  in terms of a parameter , then;

First we find . We are given that;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

If we define , then derivative of  is;

Rule for differentiation of  is;

Rule for differentiation of  is:

Similarly, we find , when we are given that;

Rule for differentiation of  is:

Rule for differentiation of  is:

If we define , then derivative of  is;

Rule for differentiation of  is;

Now we can find .

Now we can apply double angle formulae to simplify this expression.

Therefore;

We have the trigonometric identity;

So ;

We know that    provided that , therefore;

ii.

We are required to find the equation of tangent to the curve at the point where .

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

We need coordinates of the point on the curve (and tangent).

We are given equation of the curve parametrically;

We substitute desired value of parameter  in both equations of  and  to find the  coordinates of the point on the curve (and tangent).

Hence, coordinates of the point on the curve (and tangent) are .

Next we need to find slope of tangent at in order to write its equation.

The slope of a curve  at a particular point is equal to the slope of the tangent to the curve at the  same point;

Therefore, if we can find slope of the curve C at point  then we can find slope of the  tangent to the curve at this point.

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that  particular point.

Gradient (slope)  of the curve  at a particular point  can be found by substituting x- coordinates of that point in the expression for gradient of the curve;

We have found in (i) that;

Therefore;

except where  or undefined

Hence;

With coordinates of a point on the tangent  and its slope  in hand, we can write equation of the tangent.

Point-Slope form of the equation of the line is;

iii.

We are required to find the coordinates of the points on the curve where the tangent is parallel to  the x-axis, for the part of the curve where .

If two lines are parallel to each other, then their slopes  and  are equal;

We know that x-axis is horizontal and has ZERO slope, therefore;

The slope of a curve  at a particular point is equal to the slope of the tangent to the curve at the  same point;

Therefore, gradient of the curve at the point where tangent is parallel to x-axis must be ZERO as  well.

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that  particular point.

Gradient (slope)  of the curve  at a particular point  can be found by substituting x- coordinates of that point in the expression for gradient of the curve;

We have found in (i) that;

Therefore;

Taking reciprocal of both sides;

We know that  at;

 Properties of Domain Range Periodicity Odd/Even Translation/ Symmetry

We utilize the periodicity/symmetry property of   to find other solutions (roots) of :

Therefore;

For;

Only following solutions (roots) are within the given interval

Hence, the curve and tangent will have gradient equal to slope of x-axis when   and

We can find coordinates of the points on the curve when  and  by substituting these values in parametric equations, one by one.

 When Hence, coordinates of point are
 When Hence, coordinates of point are