Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2012  January  Q#10
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Question
Figure 2 shows a sketch of the curve C with equation
, x ≠ 0
The curve crosses the xaxis 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.
Solution
a.
We are given that equation of the curve C is;
We are also given that curve crosses the xaxis at the point A.
We are required to find coordinates of the point A.
It is evident that point A is the xintercept of the curve C.
The point at which curve (or line) intercepts xaxis, 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 .
b.
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 (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope 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;
Therefore;
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:
Therefore;
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.
PointSlope form of the equation of the line is;
c.
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, xcoordinate of point B is 8.
With xcoordinate of point of intersection of two lines (or line and the curve) at hand, we can find the ycoordinate of the point of intersection of two lines (or line and the curve) by substituting value of xcoordinate of the point of intersection in any of the two equations.
We choose equation of the curve;
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