Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2002 | Oct-Nov | (P2-9709/02) | Q#7


The equation of a curve is

     i.       Show that

   ii.       Find the coordinates of the points on the curve where the tangent is parallel to the x-axis.



We are given;

We are required to find .

To find  from an implicit equation, differentiate each term with respect to , using the chain rule to  differentiate any function  as .

We differentiate each term of the equation, on by one, with respect to x applying following rules.

Rule for differentiation of  is:

If  and  are functions of , and if , then;

Rule for differentiation of  is:

Now we can combines derivatives of all terms of the equation as;



We are given that tangent to the curve at a point is parallel to x-axis.

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

Since slope of x-axis is ZERO, therefore, slope of tangent to the curve is also ZERO.

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, slope of the curve at the point where tangent meets the curve is equal to the slope of the  tangent. Hence;

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve  with respect to  is:

As demonstrated in in (i), gradient of the curve is given by;

As found above;

We can substitute this  in given equation of the curve;

With x-coordinate of a point at hand, we can find the y-coordinate of the point by substituting value  of x-coordinate of the point any of the two equations.

We substitute , one by one, in equation .

Therefore, at two points with coordinates  and  the slope of tangent to curve will be  ZERO ie tangent will be parallel to x-axis.