Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2007  June  Q#10
Question
The curve C has equation , .
The points P and Q lie on C and have xcoordinates 1 and 2 respectively.
a. Show that the length of PQ is .
b. Show that the tangents to C at P and Q are parallel.
c. Find an equation for the normal to C at P, giving your answer in the form ax + by + c = 0, where a, b and c are integers.
Solution
a.
We are given that the points P and Q lie on C and have xcoordinates 1 and 2 respectively.
We are required to find the length of PQ.
Expression to find distance between two given points and is:
For the length of PQ;
Therefore, we need coordinates of the points P and Q.
With xcoordinate of a point on the line or the curve at hand, we can find the ycoordinate of the point by substituting value of xcoordinate of the point in equation of the line or the curve.
We are given that the points P and Q lie on curve C and equation of C is;
We substitute xcoordinate of the point in equation of the curve;
For the point P x=1; 
For the point Q x=2; 










Hence coordinates of point P(1,1) and Q(2,14).
Now we can find length of PQ.
b.
We are required to show that tangent to curve at points P and Q are parallel.
If two lines are parallel to each other, then their slopes and are equal;
Therefore, we need to find the slopes of both tangents to the curve at points P and Q.
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, we need to find the slopes (gradient) of the curve at the two points P and Q.
We are given equation of the curve as;
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 C at points P(1,1) and Q(2,14).
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 above expression for gradient of the given curve as;
For gradient of the curve at points P(1,1) and Q(2,14), substitute and in derivative of the equation of the curve.
For gradient at the point P x=1; 
For gradient at the point Q x=2; 








Hence, gradient of the curve C at points P(1,1) and Q(2,14) are;
Therefore, tangents to the curve at these two points will also have same slopes and hence, would be parallel.
c.
We are required to find the equation of the normal to the curve at the point P(1,1).
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 P(1,1). 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(1,1) if we have slope of curve at the same point.
We have already found in (b) slope of the curve at point P(1,1).
Now we can find slope of normal to the curve at point P(1,1).
Now we can write equation of the normal.
PointSlope form of the equation of the line is;
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