Solution of Past Paper of CAIE Math 9709/13 Oct/Nov 2019 Question # 11

CIE/GCE/AS/Math/P1/19/Nov/13/Q#11

Question




The diagram shows part of the curve 𝑦 = (𝑥 − 1)−2 + 2, and the lines x = 1 and x = 3.
The point A on the curve has coordinates (2, 3). The normal to the curve at A crosses
the line x = 1 at B.
1
(i) Show that the normal AB has equation 𝑦 = 𝑥 + 2.
2
(ii) Find, showing all necessary working, the volume of revolution obtained when
the shaded region is rotated through 360𝑜 about the x-axis.

Solution
i.


We are required to show that line AB, which is normal to the curve at point A (2, 3), has
equation;
1
𝑦 = 𝑥+2
2

, CIE/GCE/AS/Math/P1/19/Nov/13/Q#11

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).

Therefore, to find the equation of line AB we already have coordinates of a point on the
line i.e., A(2, 3).

Now we need slope of the line AB.

We are given that line AB is normal to the curve at point A(2, 3).

If a line 𝐿 is normal to the curve 𝐶, then product of their slopes 𝑚𝐿 and 𝑚𝐶 at that point
(where line is normal to the curve) is;

𝑚𝐿 × 𝑚𝐶 = −1

𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝐶𝑢𝑟𝑣𝑒 × 𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑁𝑜𝑟𝑚𝑎𝑙 𝑡𝑜 𝑡ℎ𝑒 𝐶𝑢𝑟𝑣𝑒 = −1

𝑚𝑐 × 𝑚𝑛 = −1

Therefore, if we have the slope of the given curve at point A(2, 3) we can find slope of
the normal to the curve at this point.


Let’s first find slope of the curve.

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

𝑑𝑦
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
𝑑𝑥

We have equation of the curve given as;

𝑦 = (𝑥 − 1)−2 + 2
Therefore;
𝑑𝑦 𝑑
= ((𝑥 − 1)−2 + 2)
𝑑𝑥 𝑑𝑥

Rule for differentiation of 𝑦 = 𝑔(𝑥 ) + ℎ(𝑥) is: