# 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

Therefore;

Rule for differentiation is of

Rule for differentiation is of

Rule for differentiation is of

**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

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

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

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)

Finally;

We have found that at point P gradient of the curve is

**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

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

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

With coordinates of a point on the tangent P

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

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