Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2011 | Oct-Nov | (P2-9709/22) | Q#6

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  Question

The parametric equations of a curve are

x = 1 + 2 sin2θ , y = 4 tanθ ,


i.
Show that

   ii. Find the equation of the tangent to the curve at the point where , giving your answer in  the form y = mx + c.

Solution

     i.
 

We are given that;

We are required to show that;

If a curve is given parametrically by equations for  and  in terms of a parameter , then; 

First we find . We are given that;

Therefore;

Rule for differentiation of  is;

Similarly, we find , when we are given that;

Rule for differentiation of  is:

First we solve .

Rule for differentiation of  is:

Therefore;

Next we solve .

If we define , then derivative of  is;

If we have  and then derivative of  is;

Let , then . Therefore;

Rule for differentiation of  is:

Therefore;

And;

Rule for differentiation of  is;

Hence;

Finally;

Now we can find .

  provided that

 

   ii.
 

We are required to find the equation of tangent to the curve at the point where .

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 need coordinates of the point on the curve (and tangent).

We are given equation of the curve parametrically;

We substitute desired value of parameter in both equations of  and to find the  coordinates of the point on the curve (and tangent).

Hence, coordinates of the point on the curve (and tangent) are

Next we need to find slope of tangent at in order 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 slope of the curve C at point then we can find slope of the tangent to  the curve at this 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;

We have found in (i) that;

Therefore;

Hence;

With coordinates of a point on the tangent  and its slope in hand, we can write equation  of the tangent.

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

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