Review question

# Can we sketch these four related curves? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8427

## Solution (iii)+(iv)

Hence sketch the curves $\text{(iii)}\ y = \left|\frac{x-a}{x}\right|,\qquad \dots$

As we already have a sketch of the graph $y=\frac{x-a}{x}$, we need only consider the effect that taking the absolute value has on the graph.

Every point on the curve with a positive $y$ co-ordinate is fixed and every point with a negative $y$ co-ordinate is reflected in the $x$-axis. This gives us the following sketch.

Hence sketch the curves $\dots,\qquad \text{(iv)}\ y^2 = \frac{x^2-a^2}{x^2}.$

#### Approach 1

We start by rewriting our equation as

$$$y=\pm \sqrt{1-\frac{a^2}{x^2}}. \label{equation}$$$

Firstly, we note that if $-a<x<a$, then $1-\frac{a^2}{x^2} <0$ and so $\eqref{equation}$ has no solutions. This means that our curve only has points with $x\ge a$ or $x \le -a$. We will consider the sketch of $y=1-\frac{a^2}{x^2}$, from (ii), in this range and aim to plot the square roots. This is, for each point $(x,y)$ on this curve, we want to include the points $(x,\pm \sqrt{y})$ in our sketch of $\eqref{equation}$.

If $y<1$, then the vertical distance decreases and, if $y>1$, the vertical distance increases. Therefore, we get the following graph for the positive root $y=+\sqrt{1-\frac{a^2}{x^2}}$.

To plot the other root $y=-\sqrt{1-\frac{a^2}{x^2}}$, we can simply add the reflection in the $x$-axis.

We now need only consider the points $(\pm a,0)$ and the angles that the curves make at these points. Comparing to the graph of $y=\sqrt{x}$, we would not expect the curve to be sharp at these points. Hence, we will include the information that the curve intersects the $x$-axis at right-angles in our final sketch.

Similarly to before, the equations of the asymptotes are $y=\pm 1$ and the points of intersection are at $x=\pm a$.

The full details as to why this argument works will be touched on at university-level, but there are a number of ways that we can convince ourselves of this result none-the-less. For example, would we expect these points to be sharp if we rewrote the curve in the form $x=\pm\frac{a}{\sqrt{1-y^2}}$? Alternatively, what would we expect if we used calculus to calculate the gradient of the curve at these points? What does this tell us?

#### Approach 2

We begin by rewriting our equation as

$$$y^2 + \left(\frac{a}{x}\right)^2 = 1. \label{eq:4}$$$

As this looks similar to the equation for a circle, we will begin by considering the graph of the unit circle $y^2+x^2=1$.

To get the graph of equation $\eqref{eq:4}$, we want to swap the $x$ for $\frac{a}{x}$. This is equivalent to finding another point $x'$ such that $xx'=a$ (i.e. point $x'$ for which the geometric mean of $x$ and $x'$ are on the lines $x=\pm\sqrt{a}$). This is equivalent to inversion in the lines $x=\pm\sqrt{a}$ and so each half of the circle can be projected outwards in the $x$-direction, as shown.