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.

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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