Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2013 | January | Q#11
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Question
The curve C has equation
,
a. Find , giving each term in its simplest form.
The point P on C has x-coordinate equal to .
b. Find the equation of the tangent to C at P, giving your answer in the form y = ax + b, where a and b are constants.
The tangent to C at the point Q is parallel to the line with equation 2x − 3y + 18 = 0.
c. Find the coordinates of Q.
Solution
a.
We are given;
We are required to find .
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:
b.
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 (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 need y-coordinate of the point P.
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 that point P lies on the curve C which has equation;
We substitute x-coordinate of point P in the equation of the curve C;
Since , we do not consider
.
Hence, coordinates of point P are .
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 have already found in (a);
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.
Point-Slope form of the equation of the line is;
c.
We are given that the tangent to C at the point Q is parallel to the line with equation 2x − 3y + 18 = 0.
If two lines are parallel to each other, then their slopes and
are equal;
Therefore;
We first find slope of the line with given equation;
Slope-Intercept form of the equation of the line;
Where is the slope of the line.
We can rearrange the given equation in slope-intercept form.
Therefore;
Since this line 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 x-coordinate of point B.
With x-coordinate of a point on the line (or the curve) at hand, we can find the y-coordinate of the point by substituting value of x-coordinate of the point equation of the line (or the curve).
We have equation of the curve;
Substitute ;
Since , we do not consider
.
Hence, coordinates of the point Q are (9,-1).
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