Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | Oct-Nov | (P1-9709/13) | Q#10

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Question

A curve is such that  , where a is a positive constant. The point A(a2, 3) lies on

the  curve. Find, in terms of a,

     i.       the equation of the tangent to the curve at A, simplifying your answer,

   ii.       the equation of the curve. 

It is now given that B(16,8) also lies on the curve.

  iii.       Find the value of a and, using this value, find the distance AB.

Solution

     i.
 

We are given that;

We are required to find the equation of the tangent to the curve at A(a2, 3).

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 o a point on the tangent A(a2,3).

Therefore, we need slope of the tangent to the curve to write its equation.

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 can find the slope of the curve at point A(a2,3), same would be slope of the tangent  to the curve at this 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 are given that;

Therefore, to find slope of the curve at point A(a2,3) we substitute x-coordinates of point A, ,  in above equation.

Hence;

Therefore, now we can write equation of tangent.

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


ii.
 

We are given that;

We are required to find the equation of the curve where A(a2, 3) is a point on the curve.

We can find equation of the curve from its derivative through integration;

Therefore;

Rule for differentiation of  is:

Rule for integration of  is:

If a point   lies on the curve , we can find out value of . We substitute values of  and    in the equation obtained from integration of the derivative of the curve i.e. .

We are given that A(a2, 3) is a point on the curve. Therefore, substituting coordinates of A in the  above equation;

Hence, equation of the tangent is;

  iii.
 

We are given that B(16,8) is a point on the curve.

We are required to find the value of .

From (ii) we have equation of the curve;

Substituting coordinates of point B(16,8) in equation of the curve;

Now we have two options.

We are given that  is a positive constant, therefore, only  is the possible value.

Next we are required to find the distance AB.

Expression to find distance between two given points  and is:

We are already given coordinates of points A(a2, 3) and B(16,8)

We have found above that , therefore, A(4, 3) and B(16,8).

Hence;

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