# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01R) | Year 2014 | June | Q#11

Question

A sketch of part of the curve C with equation

, x>0

is shown in Figure.

Point A lies on C and has an x coordinate equal to 2

a.   Show that the equation of the normal to C at A is y = –2x + 7.

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

b.   Use algebra to find the coordinates of B.

Solution

a.

We are required to find equation of normal to curve C at point A(2,y).

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 x-coordinates of a point on the normal A(2,y). Therefore, we need y-coordinates of  a point on the normal A(2,y) and slope of the normal to write its equation.

First we find y-coordinate of the point A on the normal. Since this point also lies on the curve, we  can utilize equation of the curve to find y-coordinate of point A(2,y).

If a point lies on the curve (or the line), the coordinates of that point satisfy the equation of the curve  (or the line).

We are given that point A(2,y) lies on the curve which has equation;

We substitute coordinates of point A in the equation of the curve C;

Hence, coordinates of point A(2,3).

Next we find slope of the normal to the curve C.

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 can find slope of normal to the curve at point A(2,y) if we have slope of curve at the same point.

We need to find the gradient of the curve C at point A(2,y).

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 that;

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:

Therefore;

Rule for differentiation is of  is:

Rule for differentiation is of  is:

Rule for differentiation is of  is:

We need to find the gradient of the curve when x=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;

Substitute  in derivative of the equation of the curve.

Hence, gradient of the curve C at points (2,y) is;

Now we can find slope of normal to the curve at point (2,y).

Now we can write equation of the normal.

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

b.

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

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;

Equating both equations;

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 .

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;

Hence, coordinates of point B are .