# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2010 | May-Jun | (P1-9709/11) | Q#7

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**Question**

The diagram shows part of the curve , which crosses the x-axis at A and the y-axis at B. The normal to the curve at A crosses the y-axis at C.

** i. **Show that the equation of the line AC is .

** ii. **Find the length of BC.

**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).

Since we are looking for the equation of normal to the curve at point A, the point A lies on the normal. Therefore, we need coordinates of the point A and slope of the normal to the curve.

We are given that point A is the x-intercept of the given curve.

The point at which curve (or line) intercepts x-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).

For the given case;

Therefore;

Therefore;

Now we need slope of the normal to find its equation. We can find the slope of the normal to the curve at point B by finding slope 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;

Rule for differentiation of is:

Therefore;

Rule for differentiation of is:

Rule for differentiation of is:

Hence;

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;

For the given case . The slope of the curve at point ;

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 slope of the normal to the curve at ;

Now we can write the equation of the normal to the curve at point .

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

Hence;

ii.

Expression to find distance between two given points and is:

Therefore, to find length of BC we need coordinates of the points B & C. First we find coordinates of B.

We know that point B is the y-intercept of the given curve.

The point at which curve (or line) intercepts y-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).

For the given case;

Therefore;

Hence coordinates of point B are .

Now we find coordinates of point C. We know that point C is the y-intercept of normal to the curve at point A.

The point at which curve (or line) intercepts y-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).

For the given case;

Therefore;

Hence coordinates of point B are .

Hence;

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