# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2007 | January | Q#8

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Question

The curve C has equation , x > 0.

a.   Find an expression for .

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

c.   Show that an equation of the normal to C at the point P is 3y=x + 20.

The normal to C at P cuts the x-axis at the point Q.

d.   Find the length PQ, giving your answer in a simplified surd form.

Solution

a.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence  gradient of curve with respect to is: We are given equation of the curve as; Therefore; Rule for differentiation is of is:  Rule for differentiation is of is: Rule for differentiation is of is:    b.

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 equation of the curve C as; We are also given coordinates of the point P(4,8).

We substitute x-coordinate of point P in the equation of the curve C;      Since coordinates of the point P satisfy the equation of the curve C, therefore the  point lies on the curve.

c.

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;   Therefore; We can find slope of normal to the curve at point P(4,8) if we have slope of curve at  the same point.

First we find slope of the curve at point P(4,8).

The slope of a curve at a particular point is equal to the slope of the tangent to the  curve at the same point;  Therefore, if we have gradient for the curve at point P, we can find the slope of tangent to the curve at point P(4,8).

First we find slope of the curve 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; We have found in (a) expression for gradient of the curve as; Now we need gradient of the curve at point P(4,8). Therefore, we substitute ;       Hence, 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;       d.

We are required to find the length PQ.

Expression to find distance between two given points and is: Therefore, we need coordinates of both points P and Q.

We are already given P(4,8).

We need to find the coordinates of point Q.

We are given that Q is the point where normal to the curve C at point P cuts the x-axis.

In other words, point Q is the x-intercept of normal to the curve C at point P.

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).

We have already found equation of normal to the curve C at point P  in (d) as; Therefore, we substitute in tis equation;   Hence, coordinates of point Q(-20,0).

Now we can find length PQ with P(4,8) and Q(-20,0) at hand.        Since ;  