<p>The price elasticity of demand is defined as:</p>

<p><span class="math-container">$$E_P=\frac{dQ}{dP} \frac{P}{Q}$$</span></p>

<p>Although generally elasticity depends on price there is a special type of functions (isoelastic functions) for which elasticity remains the same along the whole function.</p>

<p>For example consider demand given by:</p>

<p><span class="math-container">$$P=AQ^{1/e}$$</span></p>

<p>This demand function will always have the same price elasticity <span class="math-container">$E_P=e$</span> which you can verify by first solving the function above for Q and then applying the formula mentioned above. For <span class="math-container">$|e|=1$</span> the elasticity would always be unity. This is because the elasticity does not depend on anything else for example for a linear demand function the elasticity would depend on price as for <span class="math-container">$Q=a-P$</span> the <span class="math-container">$E_P=-P/(a-P)$</span> and hence as you correctly pointed out along line the elasticity constantly changes with price.</p>

<p>When you plot the function above you will get a picture that looks like hyperbola. See the picture below. However, not all functions that describe hyperbola have such property I don’t think your teacher meant to say all hyperbolas have constant elasticity.</p>

<p><a href="https://i.sstatic.net/KM40S.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/KM40S.jpg" alt="Example from Wolfram Alpha"></a>.</p>

<p>Also the demonstration project from wolfram alpha where I got the picture has amazing utility that graphically shows how the elasticity remains same along every point of the curve. The utility even allows you to play with the parameters of the model. Here is the <a href="https://demonstrations.wolfram.com/ConstantPriceElasticityOfDemand/" rel="nofollow noreferrer">link</a> for it if you want to explore it.</p>

<p>——————————————————-</p>

<p>Per request in the comment this is how you show that the elasticity is <span class="math-container">$e$</span> for <span class="math-container">$P=AQ^{1/e}$</span></p>

<ol>
<li>Solve for <span class="math-container">$Q$</span> which gives you:</li>
</ol>

<p><span class="math-container">$$Q= \frac{1}{A} P^e$$</span></p>

<ol start="2">
<li>Apply the formula for elasticity which gives you:</li>
</ol>

<p><span class="math-container">$$E_p=e\frac{1}{A}P^{e-1} \frac{P}{Q}$$</span></p>

<ol start="3">
<li>Substitute for Q the original expression <span class="math-container">$Q= \frac{1}{A} P^e$</span></li>
</ol>

<p><span class="math-container">$$E_p=e\frac{1}{A}P^{e-1} \frac{P}{ \frac{1}{A} P^e}$$</span></p>

<ol start="4">
<li>Simplify the above expression and you will get:</li>
</ol>

<p><span class="math-container">$$E_P=e$$</span></p>

CC BY-SA 4.0
2020-05-20T14:12:22.083Z
Hey this is extremely helpful! Thanks a lot! Also can you also provide me how you got e from that function, I know derivatives, so if you can provide me with a link which shows it, it would be great!
divyam sureka
CC BY-SA 4.0
2020-05-20T14:24:34.740Z
@divyamsureka I edited my answer to show you the steps. I am glad you liked the answer if you think it answers your question consider accepting the answer. Good luck with your studies
1muflon1
CC BY-SA 4.0
2020-05-20T18:14:20.203Z
from step 3, it seems we will get P as the answer rather than e, how come?
divyam sureka
CC BY-SA 4.0
2020-05-20T18:16:03.647Z
@divyamsureka no you get e. $P^{e-1}*P= P^e$ which will get you $e*[(1/A)*P^e]/[(1/A)*P^e]$ and so everything else just cancels except for e.
1muflon1
CC BY-SA 4.0
2020-05-20T18:38:18.533Z
ohhk, understood, my bad
divyam sureka
CC BY-SA 4.0
2020-05-21T20:46:36.643Z
Some minus signs are missing.
Michael
CC BY-SA 4.0
2020-05-21T20:53:16.557Z
@Michael where exactly? I tried to double check the math and can’t find a mistake - or do you mean that e=-1? I set it to 1 since it’s customary to express elasticities in absolute value. I edited it to make it more explicit
1muflon1
CC BY-SA 4.0
2020-05-20T13:28:21.853Z
2020-05-21T20:55:33.657Z
2020-05-21T20:55:33.657Z
1muflon1
<p>My teacher said that the graph of unitary elastic demand is a parabola:
<a href="https://i.sstatic.net/NhQge.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/NhQge.png" alt="enter image description here"></a></p>

<p>But i fail to understand how in a hyperbola the percentage change of price and quantity demanded remains same. Can someone explain the same to me? First i thought that it remains the same rather in a straight line but a quick study proved that it is not the case in a line with a negative slope. But the question remains the same, how is it in a hyperbola the percentage change remains the same.</p>

<econometrics><elasticity><price><demand>
Why is the graph of unitary elastic demand a hyperbola?
divyam sureka
# Why is the graph of unitary elastic demand a hyperbola?

(Asked by: username_0 on 2020-05-20)

My teacher said that the graph of unitary elastic demand is a parabola:
[![enter image description here](https://i.sstatic.net/NhQge.png)](https://i.sstatic.net/NhQge.png)

But i fail to understand how in a hyperbola the percentage change of price and quantity demanded remains same. Can someone explain the same to me? First i thought that it remains the same rather in a straight line but a quick study proved that it is not the case in a line with a negative slope. But the question remains the same, how is it in a hyperbola the percentage change remains the same.

**Tags:** econometrics, elasticity, price, demand


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## 1 answers

### Answer by username_1 on 2020-05-20. Score: 4 | Accepted answer

The price elasticity of demand is defined as:

$$E\_P=\frac{dQ}{dP} \frac{P}{Q}$$

Although generally elasticity depends on price there is a special type of functions (isoelastic functions) for which elasticity remains the same along the whole function.

For example consider demand given by:

$$P=AQ^{1/e}$$

This demand function will always have the same price elasticity $E\_P=e$ which you can verify by first solving the function above for Q and then applying the formula mentioned above. For $|e|=1$ the elasticity would always be unity. This is because the elasticity does not depend on anything else for example for a linear demand function the elasticity would depend on price as for $Q=a-P$ the $E\_P=-P/(a-P)$ and hence as you correctly pointed out along line the elasticity constantly changes with price.

When you plot the function above you will get a picture that looks like hyperbola. See the picture below. However, not all functions that describe hyperbola have such property I don’t think your teacher meant to say all hyperbolas have constant elasticity.

[![Example from Wolfram Alpha](https://i.sstatic.net/KM40S.jpg)](https://i.sstatic.net/KM40S.jpg).

Also the demonstration project from wolfram alpha where I got the picture has amazing utility that graphically shows how the elasticity remains same along every point of the curve. The utility even allows you to play with the parameters of the model. Here is the [link](https://demonstrations.wolfram.com/ConstantPriceElasticityOfDemand/) for it if you want to explore it.

——————————————————-

Per request in the comment this is how you show that the elasticity is $e$ for $P=AQ^{1/e}$

1. Solve for $Q$ which gives you:

$$Q= \frac{1}{A} P^e$$

2. Apply the formula for elasticity which gives you:

$$E\_p=e\frac{1}{A}P^{e-1} \frac{P}{Q}$$

3. Substitute for Q the original expression $Q= \frac{1}{A} P^e$

$$E\_p=e\frac{1}{A}P^{e-1} \frac{P}{ \frac{1}{A} P^e}$$

4. Simplify the above expression and you will get:

$$E\_P=e$$

> **Comments:**
> **username_0**: Hey this is extremely helpful! Thanks a lot! Also can you also provide me how you got e from that function, I know derivatives, so if you can provide me with a link which shows it, it would be great!  
> **username_1**: @username_0 I edited my answer to show you the steps. I am glad you liked the answer if you think it answers your question consider accepting the answer. Good luck with your studies  
> **username_0**: from step 3, it seems we will get P as the answer rather than e, how come?  
> **username_1**: @username_0 no you get e. $P^{e-1}*P= P^e$ which will get you $e*[(1/A)*P^e]/[(1/A)*P^e]$ and so everything else just cancels except for e.  
> **username_0**: ohhk, understood, my bad  
> **username_2**: Some minus signs are missing.  
> **username_1**: @username_2 where exactly? I tried to double check the math and can’t find a mistake - or do you mean that e=-1? I set it to 1 since it’s customary to express elasticities in absolute value. I edited it to make it more explicit  

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