# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2014 | June | Q#10

**Question**

A curve with equation y=f(x) passes through the point (4,25).

Given that

**a. **find f(x) simplifying each term.

**b. **Find an equation of the normal to the curve at the point (4, 25).

Give your answer in the form ax + by + c = 0, where a, b and c are integers to be found.

**Solution**

**a.
**

We are given;

We are given coordinates of a point on the curve (4,25).

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 (4,25) in above equation;

Therefore, equation of the curve C is;

**b.
**

We are required to find the equation of the normal to the curve at the point (4,25).

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 coordinates of a point on the normal (4,25). Therefore, we need slope of the normal to write its equation.

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 normal to the curve at point (4,25) if we have slope of curve at the same point.

We need to find the gradient of the curve C at point P(4,25).

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;

Therefore, we need .

We are already given that;

For gradient of the curve at points (4,25), substitute in derivative of the equation of the curve.

Hence, gradient of the curve C at points (4,25) is;

Now we can find slope of normal to the curve at point (4,25).

Now we can write equation of the normal.

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

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