# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2008 | June | Q#9

**Question**

The curve C has equation , where k is a constant.

**a. **Find .

Point A with x-coordinate lies on C. The tangent to C at A is parallel to the line with equation .

Find

**b. **The value of k.

**c. **The value of y-coordinate of A.

**Solution**

**a.
**

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:

We are given equation of the curve as;

Therefore;

Rule for differentiation is of is:

Rule for differentiation is of is:

Rule for differentiation is of is:

**b.
**

We are given that the line with equation is parallel to tangent to the curve C at point A.

If two lines are parallel to each other, then their slopes and are equal;

Therefore;

Now we find slope of the line in order to find slope of the tangent of curve C at point A.

First we find slope of the line with equation .

Slope-Intercept form of the equation of the line;

Where is the slope of the line.

Therefore, we can rearrange the given equation of line in this slope-intercept form.

Hence, slope of the given line is;

Therefore, slope of tangent to the curve C at point A is;

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, slope of the line with equation is equal to gradient of the curve C at point A.

Next we need gradient of the curve at point A.

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) expression for gradient of the curve as;

Since we know that at point A both gradient of curve C and slope of tangent are equal;

We know that x-coordinate of point A is and same point also lies on curve C because line is tangent to the curve C at point A. Therefore, we can substitute in above equation.

**c.
**

If a point lies on the curve (or the line), the coordinates of that point satisfy the equation of the curve (or the line).

Be careful! The point A lies on the curve and the tangent but NOT on the line which is parallel to tangent to the curve.

We are given equation of the curve;

We have found in (b) that . Therefore;

Therefore, substituting x-coordinate of point A in this equation we can get y-coordinate of point A.

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