# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2015 | June | Q#10

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**Question**

A curve with equation y=f(x) passes through the point (4,9).

Given that

, x > 0

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

Point P lies on the curve.

The normal to the curve at P is parallel to the line 2y + x = 0

**b. **Find x coordinate of P.

**Solution**

**a.
**

We are given;

We are given coordinates of a point on the curve (4,9).

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 (4,9) in above equation;

Therefore, equation of the curve C is;

**b.
**

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 slopeof 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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