Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2018 | Oct-Nov | (P1-9709/12) | Q#10

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Question

The equation of a curve is  and the equation of a line is , where k is a constant.


i.       
Find the set of values of k for which the line does not meet the curve.

In the case where k = 15, the curve intersects the line at points A and B.


ii.       
Find the coordinates of A and B.


iii.       
Find the equation of the perpendicular bisector of the line joining A and B.

Solution


i.
 

We are given equation of the curve and the line as follows;

We can write equation of the line in slope-intercept form.

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 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 k 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 12 & -12.

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 and curve do not intersect for;

   ii.
 

We are given equation of the curve and the line as follows;

We are given that ;

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;

Now we have two options.

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

Corresponding values of y coordinate can be found by substituting values of x in any of the two  equation i.e either equation of the line or equation of the curve.

We choose equation of the line;

Hence, line and the curve intersect at points;

 

iii.
 

We are required to write equation of perpendicular bisector of AB with points and .

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).

Since we are required to find the equation of perpendicular bisector, we need coordinates of a point  on it and we know that point is mid-point of AB.

Let’s find coordinates of mid-point (M) of AB.

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

Therefore, coordinates of M mid-point of AB with points A(4,11) and B(1,14);

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

Hence, coordinates of mid-point M are (,).

Next, we need slope of perpendicular bisector of AB to write its equation.

If two lines are perpendicular (normal) to each other, then product of their slopes and  is;

Therefore, if we can find slope of perpendicular bisector of AB if we have slope of line AB.

Let’s find the slope of line AB.

Expression for slope of a line joining points and ;

Therefore, slope of line AB having points A(4,11) and B(1,14);

Now we can find slope of perpendicular bisector of AB.

Finally, with coordinates of point M(,) on perpendicular bisector of AB and its slope  we can write equation of perpendicular bisector of AB.

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

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