Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2012 | January | Q#10


Figure 2 shows a sketch of the curve C with equation

 , x ≠ 0

The curve crosses the x-axis at the point A.

a.   Find the coordinates of A.

b.   Show that the equation of the normal to C at A can be written as 2x+8y−1=0

The normal to C at A meets C again at the point B, as shown in Figure 2.

c.   Find the coordinates of B.



We are given that equation of the curve C is;

We are also given that curve crosses the x-axis at the point A.

We are required to find coordinates of the point A.

It is evident that point A is the x-intercept of the curve C.

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, we substitute y=0 in equation of the curve;

Hence, coordinates of the point .


We are required to find the equation of the normal to the curve at the point .

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 already have coordinates of a point on the normal . Therefore, we need slope of the  normal to write its equation.

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;


We can find slope of normal to the curve at point  if we have slope of curve at the same  point.

We need to find the gradient of the curve C at point .

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;

Therefore, we need .

We are given;

We are required to find .

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


Rule for differentiation is of  is:

Rule for differentiation is of  is:

Rule for differentiation is of  is:

For gradient of the curve at points , substitute  in derivative of the equation of the  curve.

Hence, gradient of the curve C at points  is;

Now we can find slope of normal to the curve at point P(4,-8).

Now we can write equation of the normal.

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


We are given that the normal to C at A meets C again at the point B, as shown in Figure 2. 

Therefore, normal and the curve intersect each other at point B and we are required to find the  coordinates of B.

If two lines (or a line and a curve) intersect each other at a point then that point lies on both lines i.e.  coordinates of that point have same values on both lines (or on the line and the curve).  Therefore, we can equate  coordinates of both lines i.e. equate equations of both the lines (or the  line and the curve).

Equation of the normal is;

Equation of the curve is;

Substitute this value of y in equation of the normal;

Now we have two options.

Two values of x indicate that there are two intersection points.

We know that   corresponds to point A, therefore, x-coordinate of point B is -8. 

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 choose equation of the curve;