Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2017  FebMar  (P19709/12)  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 (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 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.
PointSlope 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 (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 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.
PointSlope 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 xcoordinate of point of intersection of two lines (or line and the curve) at hand, we can find the ycoordinate of the point of intersection of two lines (or line and the curve) by substituting value of xcoordinate 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 (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 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.
PointSlope 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 xcoordinate of point of intersection of two lines (or line and the curve) at hand, we can find the ycoordinate of the point of intersection of two lines (or line and the curve) by substituting value of xcoordinate 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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