# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2010 | May-Jun | (P1-9709/13) | Q#5

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**Question**

The equation of a curve is such that . Given that the curve passes through the point , find

** i. **the equation of the normal to the curve at P

** ii. **the equation of the curve.

**Solution**

** i.
**

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 have coordinates of one point on the normal to the curve at point . Therefore, we need slope of the normal to find 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;

Hence, we can find slope of the normal to the curve at point P from the slope (gradient) of the curve at the same point.

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

For the given case, we have;

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, slope of the curve at point .

Since;

Finally, we can write equation of the normal to the curve at point with

.

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

** ii.
**

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

For the given case, we have;

We can rearrange the expression as;

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. .

Since point lies on the curve;

Therefore equation of the curve is;

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