# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2005 | January | Q#9

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**Question**

The gradient of the curve C is given by

The point P(1,4) lies on C.

**a) **Find an equation of the normal to C at P.

**b) **Find an equation for the curve C in the form .

**c) **Using , show that there is no point on C which the tangent is parallel to the line .

**Solution**

**a) **

We are required to find the equation of the normal to the curve at the point P(1,4).

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 P(1,4). 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 P(1,4) if we have slope of curve at the same point.

First we find slope of the curve at point P(1,4).

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 gradient for the curve at point A, we can find the slope of tangent to the curve at point P(1,4).

First we find slope of the curve at point P(1,4).

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 gradient of the curve as;

Now we need gradient of the curve at point P(1,4). Therefore, we substitute ;

Hence,

Now we can find slope of normal to the curve at point P(1,4).

Now we can write equation of the normal.

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

**b) **

We are required to find the equation of the curve C.

We are required to find the equation of the curve which has point P(1,4).

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

We are given that;

Therefore;

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, we substitute coordinates P(1,4) in above equation.

Hence, equation of the curve is;

**c)
**

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

Therefore, if tangent to the curve at a point is parallel to the line with equation its slope must be equal to the slope of this given line.

Slope-Intercept form of the equation of the line;

Where is the slope of the line.

We are given equation of the line as;

We cab rearrange the equation as slope-intercept form;

Hence, slope of the line is -2.

The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;

If there is no such point on the curve such that tangent to curve at that point has slope -2, then curve does not have gradient equal to -2 at any point.

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

We are given the gradient of the curve C is given by

It is evident that but slope of the given line is negative, hence, gradient of the curve will never be equal to that of line ie there does not exist such point on the curve where curve has gradient -2.

By extension, there is no such point on the curve where tangent to the curve will have same slope as that of given line ie -2.

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