# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2005 | Oct-Nov | (P2-9709/02) | Q#4

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**Question**

The equation of the curve is .

** i. **Show that

** ii. **Find the equation of the tangent to the curve at the point (2, 4), giving your answer in the form ax+by=c.

**Solution**

** i.
**

We are given that;

Therefore;

Rule for differentiation of is:

If and are functions of , and if , then;

If , then;

Rule for differentiation of is:

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

** ii.
**

We are required to find the equation of tangent to the curve at the point (2,4).

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 the point on the curve (and tangent) as (2,4).

Next we need to find slope of tangent 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 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 equation of the curve implicitly;

We have found in (i) that;

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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