Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2011 | Oct-Nov | (P1-9709/13) | Q#7

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Question


i.   
A straight line passes through the point (2, 0) and has gradient . Write down the equation of the line.

ii.  Find the two values of  for which the line is a tangent to the curve . For each value of , find the coordinates of the point where the line touches the curve.

  iii.  Express  in the form  and hence, or otherwise, write down the  coordinates of the minimum point on the curve.

Solution


i.
 

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

For the given case where point is (2, 0) and has gradient ;


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;

Equation of the curve is;

Equating both equations;

For a quadratic equation , the expression for solution is;

Where  is called discriminant.

If , the equation will have two roots.

If , the equation will have two identical/repeated roots.

If , the equation will have no roots.

For the given case;

The discriminant is as follows;

Two values of  indicate that there are two intersection points.

For 

Equation of line from (i) is as follows;

Substitute  and ;

Hence coordinates of one point are .

For 

Equation of line from (i) is as follows;

Substitute  and ;

Hence coordinates of one point are .


iii.
 

We have the expression;

We use method of “completing square” to obtain the desired form. We can rewrite the expression as;

Next we complete the square for the terms which involve .

We have the algebraic formula;

For the given case we can compare the given terms with the formula as below;

Therefore we can deduce that;

Hence we can write;

To complete the square we can add and subtract the deduced value of ;

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.

We recognize that given curve , is a parabola opening upwards.

Vertex form of a quadratic equation is;

The given curve  , as demonstrated above can be written in vertex form as;

Coordinates of the vertex are . Since this is a parabola opening upwards the vertex is the minimum point on the graph. Here y-coordinate of vertex represents least value of  and x-coordinate of vertex represents corresponding value of .

For the given case, vertex or minimum point has coordinates is .

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