Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2009  June  Q#11
Question
The curve C has equation
The point P has coordinates (2, 7).
a) Show that P lies on C.
b) Find the equation of the tangent to C at P, giving your answer in the form y=mx+c, where m and c are constants.
The point Q also lies on C.
Given that the tangent to C at Q is perpendicular to the tangent to C at P,
c) show that the xcoordinate of Q is
Solution
a)
If a point lies on the curve (or the line), the coordinates of that point satisfy the equation of the curve (or the line).
We are given equation of the curve C as;
We are also given coordinates of the point P(2,7).
We substitute xcoordinate of point P in the equation of the curve C;
Since coordinates of the point P satisfy the equation of the curve C, therefore the point lies on the curve.
b)
We are required to find the equation of tangent to the curve C at point P(2,7).
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 as P(2,7).
We need to find slope of tangent at in order 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 C at point P(2,7) then we can find slope of the tangent to the curve at this point.
We need to find the gradient of the curve C at point P(2,7).
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;
We are given equation of the curve as;
Therefore;
Rule for differentiation is of is:
Rule for differentiation is of is:
Rule for differentiation is of is:
For gradient of the curve at point P(2,7), substitute in derivative of the equation of the curve.
Therefore;
With coordinates of a point on the tangent P(2,7) and its slope in hand, we can write equation of the tangent.
PointSlope form of the equation of the line is;
c)
We are given that tangent to the curve C at point Q is perpendicular to the tangent to the curve C at point P.
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;
Slope of Tangent at point P X Slope of tangent at point
Q=1
We have found in (b) that;
Therefore;
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
We have from (b) that;
If we have coordinates of a point on the curve, we can find gradient of the curve at that point by substituting coordinates of this point in expression of the derivative of equation of the curve.
Conversely, if we have gradient of the curve at particular, we can equate expression of the derivative of equation of the curve with the given gradient of curve at that point to find the coordinates of that point.
Therefore;
For a quadratic equation , the expression for solution is;
Therefore;
Since ;
Now we have two options.




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