# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2006 | May-Jun | (P1-9709/01) | Q#9

Question

A curve is such that  and P (1, 8) is a point on the curve.

i.
The normal to the curve at the point P meets the coordinate axes at Q and at R. Find the coordinates of the mid-point of QR.

ii.   Find the equation of the curve.

Solution

i.

To find the mid-point of a line we must have the coordinates of the end-points of the line.

Expressions for coordinates of mid-point of a line joining points  and;

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

So first we need to find the coordinates of points Q & R.

We know that Q & R are the coordinate axes intercepts of the normal to the curve at point P.

Therefore, to find the coordinate axes intercepts of a line (the normal to the curve at point P) we need equation of the line (the normal to the curve at point P).

To find the equation of the normal to the curve at point P;

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).

The required coordinates of the point are P (1, 8), hence we only need to find the slope of the normal to the curve at P.

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 we need slope of the curve at point P;

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

.

In the given case;

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;

Coordinates of point P (1, 8);

Hence, slope of the normal to the curve at point P;

Now we can write equation of the normal to the curve at P(1,8);

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

Now we find the coordinate axes intercepts of the normal to the curve.

The point  at which curve (or line) intercepts x-axis, the value of . So we can find the value of  coordinate by substituting  in the equation of the curve (or line).

Therefore;

Similarly;

The point  at which curve (or line) intercepts y-axis, the value of . So we can find the value of  coordinate by substituting  in the equation of the curve (or line).

Now we can find the coordinates of mid-point of QR;

Expressions for coordinates of mid-point of a line joining points  and;

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

ii.

We can find equation of the curve from its derivative through integration;

In the given case;

Rule for integration of  is:

We can rewrite the equation;

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 have that point P (1, 8) on the curve;

Hence equation of the curve;