2010 Paper 2 Q5 - coord geometry

The curve y = 5 – e^2x intersects the coordinate axes at the points A and B.

(i) Given that the line AB passes through the point with the coordinates (ln 5, k), find the value of k.

(ii) In order to solve the equation x = ln √(9-x), a graph of a suitable straight line in drawn on the same set of axes as the graph of y = 5 – e^2x.  Find the equation of the straight line.

At first thought, we might get stumped by the second question. That might paralyse your thinking for the first one. But, in all problems, we start from where we are and see how to get to where we want to go. So we don’t have to really worry about it now. But we need to look at solving the first one first. So let’s get that one started…

(i)

Since A and B are points on the x- and y-axes, we will find the coordinates of A and B.

At x-axis, y = 0

So,  0 = 5 – e^2x

e^2x = 5

2x = ln 5

x = ½ln 5

Hence, A(½ln 5, 0)

At the y-axis, x = 0

So, y = 5

Hence, B(0, 5)

Gradient of AB is

Therefore, the equation of AB is  y = (10/ln 5)x + c

When x = ½ln 5, y = 0.  Therefore, c = 5

So, the equation of AB is  ln5y = 10x + 5ln5

Hence, when x = ln 5,   ln5y = 10ln5 + 5ln5

Dividing throughout by ln5, we get   y = 15

Therefore k = 15.

(ii)  Now, let’s see what we can do about the next section.

The thing is, when you want to solve the equation of x = ln √(9-x) by drawing a straight line through the curve y = 5 – e^2x, what that means is we have to find a straight line y = mx + c, such that when

                mx + c = 5 – e^2x,

the resulting equation can be simplified to give x = ln √(9-x).

So, let us work with the above equation then.

                mx + c = 5 – e^2x

                e^2x = (5-c) – mx

                e^x = √[(5-c) – mx]

                x = ln√[(5 – c) – mx]

Equating this with x = ln √(9-x), we can see that

5 – c = 9

c = -4

m = 1

Hence, the equation of the straight line to be drawn on the same graph of y = 5 – e^2x is

                y = x – 4