# Simple equation.

1. Jan 20, 2007

### ritwik06

1. The problem statement, all variables and given/known data
For the graph of equation (y^2)=a(x^2), where a is a positive constant, whic of the following is true?
1) y increases linearly with x
2) y increases as the square of x
3) y increases as y^1/2
4) If x increases, y can increase or decrease.

2. Relevant equations
y is directly proportional to x. as a is a constant, the first option would be right!

3. The attempt at a solution

Is my approach right?

2. Jan 20, 2007

### arildno

Not all, I'm afraid. Rewrite your equation as:
$$(y-(\sqrt{a})x)(y+(\sqrt{a})x)=0$$
Now, two questions:
1. Why is this rewriting permissible?
2. What must you conclude from it?

Last edited: Jan 20, 2007
3. Jan 20, 2007

### ritwik06

Ans 1 because the difference of two squares can be factorised as suc:
a^2-b^2=(a+b)(a-b)

Ans 2. I dont know what can be concluded! Please be clear!

4. Jan 20, 2007

### arildno

Well, if two numbers multiplied together is zero, what can we say about those numbers?

5. Jan 20, 2007

### ritwik06

That any one of them is equal to zero or both are zero.

6. Jan 20, 2007

### arildno

So,
EITHER:
$$y=\sqrt{a}x$$
OR
$$y=-\sqrt{a}x$$

Given that knowledge, which statement is true?

7. Jan 20, 2007

### ritwik06

The fourth option???

8. Jan 20, 2007

### arildno

Yup

9. Jan 20, 2007

### ritwik06

but sir. I agree totally with you. But if you look at it straight away, it seems that y is directly proprtional to x. hence they will have a linear relationship. How will you criticise this point? Thanks for the effort!!!!

10. Jan 20, 2007

### arildno

Three flaws:

1. As for direct proportionality (i.e, ratio constant), remember that the proportionality constant in such a relation can be positive or negative, so that from mere direct proportionality, you cannot conclude that y will increase with x, it could equally well decrease with x.

2. If we assume that the y-coordinate is to be regarded as a FUNCTION of the x-coordinate (so that we limit ourselves to only a part of the permissible solution pairs of the original equation), we might have a discontinuous function relationship in that in one x-region, the function values followed the upper line, and in an adjoining x-region, the function values follow the lower line.
In this case, we see that the ratio x/y is NOT constant, that is direct proportionality do not hold between x and y

3. There is no reason why we should assume that there exist any function relating x-values to y-values. Indeed, the solution set to the equation do not permit a function construction in general.

To be sure, if we limit ourselves to just a part of the solution set, then we may, indeed, have some functional relation between the picked (x,y)-values.
But that is a totally different issue.

11. Jan 21, 2007

### HallsofIvy

Staff Emeritus
This is really only echoing arildno but:

I would "criticise" that point by saying "Yes, it true- but off the point." The answer did not say "linearly related", it said "increases linearly". That was arildno's point.

12. Jan 22, 2007

### ritwik06

If I read the question again, it says that a is "positive constant". Are you stressing on the point that the squares of a negative number is always positive??

Whats the difference between "inearly related" and "increases linearly"?

13. Jan 22, 2007

### HallsofIvy

Staff Emeritus
Yes, and so with y2= ax2 we can have either
$y= \sqrt{a}x$ or $y= -\sqrt{a}x$

In both $y= \sqrt{a}x$ and $y= -\sqrt{a}x$ y and x are "linearly related". In the first, y "increases linearly" with x but in the second y "decreases linearly" with x.

14. Jan 22, 2007

### ritwik06

Thanks a lot for the efforts.