Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2011 | January | Q#11



The curve C has equation

 , x>0

a.   Find

b.   Show that the point P(4,−8) lies on C.

c.   Find an equation of the normal to C at the point P, giving your answer in the form ax + by + c = 0  , where a, b and c are integers.



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:


We are required to show that the point P (4, -8) lies on curve .

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 have found in (a) that equation of curve  is;

We substitute x-coordinate of point P in the equation of the curve  is;

Since coordinates of the point P satisfy the equation of the curve , therefore the point lies on the  curve.



We are required to find the equation of the normal to the curve at the point P(4,-8).

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 P(4,-8). 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 P(4,-8) if we have slope of curve at the same point.

We need to find the gradient of the curve C at point P(4,-8).

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 .

From (a) we already have found that;

For gradient of the curve at points P(4,-8), substitute  in derivative of the equation of the curve. 

Hence, gradient of the curve C at points P(4,-8) 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;