Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2017 | Oct-Nov | (P1-9709/12) | Q#7

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Question

Points A and B lie on the curve . Point A has coordinates (4,7) and B is the  stationary point of the curve. The equation of a line L is , where m is a constant.

                            i.       In the case where L passes through the mid-point of AB, find the value of m.

                          ii.       Find the set of values of m for which L does not meet the curve.

Solution


i.
 

We are given equation of line L as follows;

It is evident that it is in slope-intercept form.

Slope-Intercept form of the equation of the line;

Where  is the slope of the line.

We are required to find the value of m when line L passes through mid-point of AB. 

Therefore, if we can write equation of the line L passing through mid-point of AB  then  we can compare this with given equation of line L to find m.

Let’s find equation of line L.

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 do not have coordinates of a point on the line L but we know that its slope is m.

Let’s find coordinates of a point on the line L.

We are given that line L passes through the mid-point of AB.

Therefore, we need to find coordinates of mid-point of AB which lies on line L.

To find the mid-point of a line we must have the coordinates of the end-points of the line.

Expressions for coordinates of mid-point of a line joining points  and;

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

We have coordinates of point A only given as (4,7).

We are given that other point, B, is a stationary point on the curve with equation given as;

We can find coordinates of a stationary point on the curve as follows.

Coordinates of stationary point on the curve  can be found from the derivative of equation of the  curve by equating it with ZERO. This results in value of x-coordinate of the stationary point  on the curve.

Therefore we need to find derivative of the equation of the curve first.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve  with respect to  is:

Hence;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Now we equate this derivative with zero to find the x-coordinate of point B which is a stationary point on the given curve.

One possible values of  implies that there is only one stationary point on the curve at this value of .

To find y-coordinate of the stationary point  on the curve, we substitute value of x-coordinate  of the stationary point  on the curve (found by equating derivative of equation of the curve  with ZERO) in the equation of the curve.  

We have equation of the curve as;

Substituting ;

Hence, coordinates of point B are (2,3). 

Now we can find the coordinates of mid-point of AB with coordinates A(4,7) and B(2,3).

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

Hence coordinates of mid-point of AB which lies on line L are (3,5).

Now we can write equation of line L.

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

We are given that slope of line L is m. hence;

Comparing this equation with the given equation  of line L, we can write that;


ii.
 

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).

Equation of the line is given as;

Equation of the curve is;

Equating both equations;

Therefore, if given line L and the given curve do not intersect, there should be no solution of this  equation.

It is evident that it is a quadratic equation.

Standard form of quadratic equation is;

Expression for discriminant of a quadratic equation is;

If   ; Quadratic equation has two real roots.

If   ; Quadratic equation has no real roots.

If   ; Quadratic equation has one real root/two equal roots.

Therefore, for the given equation;

To find the set of values of m for which , we solve the following equation to find  critical values of ;

Now we have two options;

Hence the critical points on the curve for the given condition are 2 & -10.

Standard form of quadratic equation is;

The graph of quadratic equation is a parabola. If  (‘a’ is positive) then parabola opens upwards  and its vertex is the minimum point on the graph.
If
 (‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the  graph.

It is evident that  is an upward opening parabola.

Therefore conditions for  are;

Hence, line L and curve do not intersect for;

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