Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663A/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 (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).
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.
Point-Slope 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;
Slope-Intercept 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 x-coordinate of point Q.
Now we have two options.
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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 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 ;
Hence, coordinates of the point Q are (,
).
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