# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2016 | June | Q#11

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**Question**

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

**a. **Find .

The point P, where x=-2, lies on C.

The tangent to C at the point P is parallel to the line with equation 2y – 17x – 1=0.

Find

**b. **the value of k.

**c. **the value of y-coordinate of P.

**d. **the equation of the tangent to C at P, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

**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 value of k.

We have found from (a) that;

Therefore, if we have slope (gradient) of the curve at some given point, we can find the value of k.

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 have slope of tangent to curve at point P, we can find gradient of the curve at this point.

We are given that tangent to C at the point P is parallel to the line with equation 2y – 17x – 1 = 0.

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

Therefore, we can find the slope of tangent to curve at point P by finding slope of the line parallel to tangent.

Slope-Intercept form of the equation of the line;

Where is the slope of the line.

Therefore, can rearrange the following given equation of line, parallel to tangent, in slope intercept form.

Therefore, slope of the line parallel to tangent to the curve at point P is .

Hence;

Now we can also find gradient of the curve at point P.

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;

Finally;

We have found that at point P gradient of the curve is and we are also given that at point P; x= –2, therefore;

**c. **

We are required to find y-coordinate of point P.

We are given that x-coordinate of point P is -2.

We are given that point P lies on the curve C which has equation;

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

Therefore, we substitute x=-2 in equation of the curve;

We have also found in (b) that ;

**d. **

We are required to find the equation of tangent to the curve C at point P where x=-2.

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).

We already have found coordinates of a point on the tangent P.

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 P then we can find slope of the tangent to the curve at this point.

We already have found slope of tangent to the curve C at point P.

With coordinates of a point on the tangent P and its slope in hand, we can write equation of the tangent.

Point-Slope form of the equation of the line is;

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