# 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

Therefore;

Rule for differentiation is of

Rule for differentiation is of

Rule for differentiation is of

**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

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

Therefore, if we can find slope of the curve C at 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)

We have already found in (a);

For gradient of the curve at point

Therefore;

With coordinates of a point on 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

Therefore;

We first find slope of the line with given equation;

Slope-Intercept form of the equation of the line;

Where

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

Therefore;

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.

Gradient (slope)

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

Hence, coordinates of the point Q are (9,-1).

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