Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2007 | June | Q#10

Question

The curve C has equation , .

The points P and Q lie on C and have x-coordinates 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 x-coordinates 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 x-coordinate of a point on the line or the curve at hand, we can find the y-coordinate of the  point by substituting value of x-coordinate 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 x-coordinate 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 (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(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.

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

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