Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2017 | Feb-Mar | (P1-9709/12) | Q#9

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Question

The point A(2,2) lies on the curve .

                    i.       Find the equation of the tangent to the curve at A.

The normal to the curve at A intersects the curve again at B.

                  ii.       Find the coordinates of B.

The tangents at A and B intersect each other at C.

                 iii.        Find the coordinates of C.

Solution

     i.
 

We are required to find the equation of the tangent to the curve at A(2,2). 

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 tangent A(2,2). However, we need to find slope of  the tangent 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, we can find the slope of tangent from gradient of the curve at point A(2,2).

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that  particular point.

We are given equation of the curve as;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

This gives the expression for gradient of the curve.

Now we need to find gradient of the curve at point A(2,2).

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, substituting  in expression of the gradient of the curve;

Hence gradient of the curve at point A is , and therefore, slope of the tangent to the curve at  point A;

Now we can write equation of the tangent as follows.

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


ii.
 

We are required to find the coordinates of point B which is intersection point of the curve and the  normal to the curve at point A(2,2).

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).

Therefore, we need equations of both the curve and the normal to the curve at point A(2,2).

We are already given equation of the curve as;

Now we need to find equation of normal to the curve at point A(2,2).

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 A(2,2).
However, we need to find 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 the slope of normal from gradient of the curve at point A(2,2). 

We have already found in (i) gradient of the curve at point A as , and therefore, slope of the  normal to the curve at point A;

Now we can write equation of the normal as follows.

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

Now when we have equations of both the curve and the normal to the curve at point A(2,2), we can  equate these two equations to find the coordinates of point of intersection.

Now we have two options.

Two values of x indicate that there are two intersection points.

It is evident that  represents point A(2,2) where line is normal to the curve and  represents point B where normal again meets the curve.

With x-coordinate of point of intersection of two lines (or line and the curve) at hand, we can find the  y-coordinate of the point of intersection of two lines (or line and the curve) by substituting value  of x-coordinate of the point of intersection in any of the two equations;

We choose equation of normal;

Therefore, we substitute ;

Hence, coordinates of point B are .


iii.
 

We are required to find the coordinates of point C which is intersection point of the tangent to the  curve at point A(2,2) and tangent to the curve at point 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).

Therefore, we need equations of both the tangents to the curve at point A and B.

We already have found equation of tangent to the curve at point A(2,2) as;

However, we need to find equation of tangent to the curve at point B.

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 tangent B. However, we need to find slope of  the tangent 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, we can find the slope of tangent from gradient of the curve at point B

We have already found in (i) expression for the gradient of the curve;

Now we need to find gradient of the curve at point B.

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, substituting  in expression of the gradient of the curve;

Hence gradient of the curve at point B is , and therefore, slope of the tangent to the curve  at point B;

Now we can write equation of the tangent as follows.

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

Equation of tangent at A is;

Equation of tangent at B is;

Equating both equations;

Single value of x indicates that there is only one intersection point.

With x-coordinate of point of intersection of two lines (or line and the curve) at hand, we can find the  y-coordinate of the point of intersection of two lines (or line and the curve) by substituting value  of x-coordinate of the point of intersection in any of the two equations;

We choose equation of tangent;

Therefore, we substitute ;

Hence, coordinates of point C are .

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