# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2017 | June | Q#7

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**Question**

The curve C has equation y=f(x), x>0, where

Given tht the point P(4,-8) lies on the curve C;

**a. **find the equation of the tangent to C at P, giving your answer in the form y = mx + c, where m and c are constants.

**a. **find f(x), giving each term in its simplest form.

**Solution**

**a.
**

We are required to find the equation of tangent to the curve C at point P(4,-8).

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(4,-8).

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(4,-8) then we can find slope of the tangent to the curve at this 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 are given that;

Therefore;

We know that at point P(4,-8) : x= 4, therefore;

Hence;

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;

**b. **

We are given;

We are given coordinates of a point on the curve P(4,-8).

We are required to find the equation of y in terms of x ie f(x).

We can find equation of the curve from its derivative through integration;

Therefore,

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

If a point lies on the curve , we can find out value of . We substitute values of and in the equation obtained from integration of the derivative of the curve i.e. .

Therefore, substituting the coordinates of point P(4,-8) in above equation;

Therefore, equation of the curve C is;

**c.
**

We are given that point P lies on the curve and the normal to the curve at P is parallel to the line 2y + x = 0.

We are required to find the x coordinate of P.

We can find the x-coordinates of P through its gradient.

If we can find the gradient o the curve at point P, we can equate it with given expression of derivative (gradient) of the curve to find x-coordinates of point P.

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

Therefore, we need gradient of the curve at point P.

We are given that normal to the curve at point P has equation 2y + x = 0.

If a line is normal to the curve , then product of their slopes and at that point (where line is normal to the curve) is;

Therefore;

We can find slope of the curve at point P if we can find slope of the normal to the curve at this point.

We are given equation of the normal to the curve at point P.

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 as follows.

Hence, slope of the normal to the curve at point P is .

Now we can find slope of the curve at point P.

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;

Therefore;

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