# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2005 | Oct-Nov | (P1-9709/01) | Q#10

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Question

A curve is such that and (1, 4) is a point on the curve.

i.
Find the equation of the curve.

ii.        A line with gradient is a normal to the curve. Find the equation of this normal, giving your answer in the form .

iii.       Find the area of the region enclosed by the curve, the x-axis and the lines x = 1 and x = 2.

Solution

i.

We can find equation of the curve from its derivative through integration;  In this case; Therefore;  We can rewrite the expression as; Rule for integration of is: Therefore;      If a point lies on the curve , we can find out value of . We substitute values of and in the equation obtained from integration of the derivative of the curve i.e. .

We know that (1,4) is a point on the curve. Hence;   Hence equation of the curve is; ii.

To find the equation of normal to the curve;

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 have the slope of the normal to the curve; Now to find the coordinates of the point on the normal we utilize the fact that normal and the curve intersect each other.

To find coordinates of that intersection point we can find slope of the curve at that point;

If a line is normal to the curve , then product of their slopes and at that point (where line is normal to the curve) is;   Therefore;   Now we can find coordinates of the point on the curve where slope is 2.

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; In the given case: Hence;       To find the y-coordinate of this point on the curve, we substitute in the equation of the curve derived in (i).     Hence point of intersection of the curve and normal to the curve is .

Now that we have both coordinates of a point on the normal and slope of the normal , we can write the equation of the normal;

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

To find the area of region under the curve , we need to integrate the curve from point to along x-axis. In this case, we have from (i); Therefore;  We can rewrite the expression as; Rule for integration of is:  Rule for integration of is: Rule for integration of is:          