# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2019 | May-Jun | (P2-9709/21) | Q#3

**Question**

Find the equation of the normal to the curve

at the point (3, 1).

**Solution**

** **We are given equation of the curve as;

We are required to find the equation of normal to the curve at the point (3, 1).

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 need coordinates of the point on the curve (and normal).

We are already given coordinates of a point on the curve (and normal) as (3, 1).

Next we need to find slope of normal at (3, 1) in order 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, if we can find slope of the curve C at point (3, 1) then we can find slope of the normal to the curve at this point.

Therefore, we first find expression for gradient of the curve.

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

Therefore;

Rule for differentiation of is:

First we differentiate .

Rule for differentiation of is:

Next we differentiate .

To find from an implicit equation, differentiate each term with respect to , using the chain rule to differentiate any function as .

We utilize Product Rule to differentiate .

If and are functions of , and if , then;

If , then;

Let and , then;

Rule for differentiation of is:

For ;

Therefore;

Next we differentiate .

Rule for differentiation of is:

Lastly,we differentiate .

Rule for differentiation of is:

Hence;

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;

Hence;

With coordinates of a point on the tangent and its slope in hand, we can write equation of the tangent.

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

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