# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | May-Jun | (P1-9709/11) | Q#8

Question

A curve has equation  and passes through the points A(1,-1) and B(4,11). At each of  the points C and D on the curve, the tangent is parallel to AB. Find the equation of the  perpendicular bisector of CD.

Solution

We are required to find the equation of perpendicular bisector of CD.

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).

Let us first find the coordinates of a point on perpendicular bisector of CD.

It is evident that mid-point  of CD is a point on the perpendicular bisector of CD.

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

Therefore, to find coordinates of  we need coordinates of both points C and D.

We know that at each of the points C and D on the curve, there is a tangent.

We are given that this tangent is parallel to AB.

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

Therefore;

Therefore, if we have slope of AB we can find slope of tangent to the curve.

Let find slope of AB.

Expression for slope of a line joining points  and ;

Therefore, slope of AB with coordinates of A(1,-1) and B(4,11);

Hence;

The slope of a curve  at a particular point is equal to the slope of the tangent to the curve at the  same point;

It is evident that slope of the tangent to the curve can be utilized to find slope of curve at that point  where tangent meets the curve.

Therefore,  will be slope of curve at two points where the same tangent meets the  curve.

Now we find the coordinates of the point(s) on the curve where slope (gradient) of the curve is  .

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that  particular point.

We first need to find the derivative of the equation of the curve in order to find coordinates of  point(s) on the curve where slope of the curve is given.

We are given equation of the curve as;

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

Rule for differentiation of  is:

Rule for differentiation of  is:

This equation gives the gradient of the curve at any point on the curve.

If we have gradient of the curve at any point, we can use this equation to find the coordinates of  that point on the curve.

We have slope of the curve at two distinct point on the curve; it is the same as the slope of  tangent to curve at those two points.

To find the coordinates of those points on the curve, we can equate the two equations of slope of  the curve;

Two values of x indicate that there are two intersection points of tangent and the curve.

These are x-coordinates of points C and D where tangent meets the curve.

With x-coordinate of point of intersection of two lines (or line and the curve) at hand, we can find  the y-coordinate of the point of intersection of two lines (or line and the curve) by substituting  value of x-coordinate of the point of intersection in any of the two equations.

We have only equation of the curve;

Now we have two options.

 For For

Hence, two points where tangent and the curve meet have coordinates  and i.e.  points C and D.

As discussed earlier, we needed coordinates of points C and D to find the coordinates of mid-point of CD; namely poin M.

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

Hence, coordinates of a point on the perpendicular bisector of CD are .

In order to write equation of the perpendicular bisector of CD, we need slope of the perpendicular  bisector of CD.

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, if we have slope of CD we can find slope of the perpendicular bisector of CD.

Let us find slope of CD.

Expression for slope of a line joining points  and ;

With coordinates of points C and D at hand;  and  we can find slope of CD.

Therefore;

Finally, with coordinates of a point on perpendicular bisector of CD  and slope of perpendicular bisector of CD  at hand, we can write equation of perpendicular bisector  of CD.

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