# 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; 