Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663A/01)  Year 2014  January  Q#10
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Question
The curve C has equation .
The point P, which lies on C, has coordinates (2, 1).
a. Show that an equation of the tangent to C at the point P is y = 3x – 5.
The point Q also lies on C.
Given that the tangent to C at Q is parallel to the tangent to C at P,
b. find the coordinates of the point Q.
Solution
a.
We are required to find the equation of tangent to the curve C at point .
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).
Next, 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 , 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 .
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;
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
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 , substitute in derivative of the equation of the curve.
Therefore;
With coordinates of a point on the tangent and its slope in hand, we can write equation of the tangent.
PointSlope form of the equation of the line is;
b.
We are given that the tangent to C at the point Q is parallel to the tangent to C at P.
If two lines are parallel to each other, then their slopes and are equal;
Therefore;
We first find slope of the tangent to C at P.
We have found equation of the tangent to C at P in (a) as;
SlopeIntercept form of the equation of the line;
Where is the slope of the line.
Therefore;
Since tangent to C at P is parallel to the tangent to the curve at point Q, the gradient of the curve at point Q must be the same.
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;
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 have found in (a) that expression for slope of the given curve is;
Therefore;
We can solve this equation to find the xcoordinate of point Q.
Now we have two options.







Two values of x indicate that there are two points with same gradient on the curve.
We already know that P(2,1), therefore, point with must be the point Q.
With xcoordinate of a point on the line (or the curve) at hand, we can find the ycoordinate of the point by substituting value of xcoordinate of the point equation of the line (or the curve).
We have equation of the curve;
Substitute ;
Hence, coordinates of the point Q are (,).
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