# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2019 | May-Jun | (P1-9709/12) | Q#9

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Question

The curve C1 has equation y = x2 4x + 7. The curve C2 has equation y2 = 4x + k, where k is a constant. The tangent to C1 at the point where x = 3 is also the tangent to C2 at the point P. Find the  value of k and the coordinates of P.

Solution

We are given equation of the curve C1 as; We are given equation of the curve C2 as; First, we need to find the equation of the tangent to the curve C1 at point x=3.

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).

First, we find coordinates of the point where tangent meets the curve i.e., x=3.

Corresponding values of y coordinate can be found by substituting value of x in equation of the  curve.

Therefore, we substitute in equation of the curve C1;    Hence, curve and tangent meet at the point which lies on both curve and the tangent.

Next, we need slope of the tangent to the curve 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, if we can find slope of the curve at a given point, then we can find slope of the tangent to the curve at that particular point.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is: We are given equation of the curve C1; Therefore; Rule for differentiation of is:  Rule for differentiation of is: Rule for differentiation of is:    To find slope of the tangent to the curve at point we need gradient of the curve at the same point.

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

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, to find the gradient at the curve at the point we substitute in;    Now we have slope of the curve as; We can find slope of the tangent to the curve at this point of the curve i.e., point ; Now, with coordinates of a point on the tangent nd slope of the tangent , we can write equation of the tangent.

Point-Slope form of the equation of the line is;    We are given that tangent to C1 at the point where x = 3 is also the tangent to C2 at the point P.

Therefore, point P is the point of intersection of the curve C2 and tangent to C1.

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 tangent is; Equation of the curve C2 is;  Equating both equations;      Since, tangent and curve intersect at a single point only, the solution of the above equation must  result in a single value of .

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 case, the discriminant of above equation must be equal to ZERO so that  one solution results for ;   Therefore;      Next, we are required to find the coordinates of point of intersection P of the curve C2 and tangent  to C1.

We have found above that it can be found through equation; We have also found that ;           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 tangent;    Hence, coordinates of the point P are .