# 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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