# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2012 | Oct-Nov | (P2-9709/22) | Q#7

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Question

The equation of a curve is i.       Show that ii.       Find the coordinates of each of the points on the curve where the tangent is parallel to the x- axis.

Solution

i.

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;           ii.

We are given that tangent to the curve at some points 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.