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Question:

The straight-line $x\cos\alpha+y\sin\alpha=p$ intersects the x & y axes at A & B respectively. Considering $\alpha$ as a variable, show that the equation of the locus that the middle point of AB is a part of is $p^2(x^2+y^2)=4x^2y^2$.

Solution:

$$x\cos\alpha+y\sin\alpha=p$$

$$\implies \frac{x}{\sec\alpha}+\frac{y}{\csc\alpha}=p$$

$$\implies \frac{x}{p\sec\alpha}+\frac{y}{p\csc\alpha}=1$$

$\therefore$ The coordinates of A & B are $(p\sec\alpha,0)$ & $(0, p\csc\alpha)$. Moreover, the midpoint of AB is $(\frac{p\sec\alpha}{2}, \frac{p\csc\alpha}{2})$. Let, $(x,y)$ is the midpoint of any straight line AB. So,

$$(x,y)=(\frac{p\sec\alpha}{2}, \frac{p\csc\alpha}{2})$$

Now,

$$x=\frac{p\sec\alpha}{2}$$

$$\implies2x\cos\alpha=p$$

$$\implies\cos\alpha=\frac{p}{2x}...(i)$$

Again,

$$y=\frac{p\csc\alpha}{2}$$

$$\implies 2y\sin\alpha=p$$

$$\implies \sin\alpha=\frac{p}{2y}...(ii)$$

$(i)^2+(ii)^2:-$

$$\frac{p^2}{4x^2}+\frac{p^2}{4y^2}=1$$

$$\implies p^2(\frac{y^2+x^2}{4x^2y^2})=1$$

$$\implies p^2(x^2+y^2)=4x^2y^2(showed)$$

Now, my problem is with the step when we square and add (i) & (ii). How do we know that we are not introducing extraneous roots? I have a bigger question. How do I know when squaring is valid or not so that I don't have to come to Math SE every time I encounter squaring in my book?

This might help you in answering my question.

Blue
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tryingtobeastoic
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    I think we can square as and when required but we must be careful while taking square root. – Lalit Tolani Sep 06 '21 at 04:30
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    Note that you are only trying to show that your points are in the curve. So all you need to show is that they satisfy the equation, which you did. You are not claiming that every point in the curve is the mid point of $A$ and $B$. So there is no problem with squaring. What you are doing is showing that the point in question satisfies the equation in question, you aren't solving an equation. – Arturo Magidin Sep 06 '21 at 04:54
  • Often stuff like this is written as though it should be plain what the next step is. In reality authors often solved it by some other means then through hindsight found a shorter way to arrive at the solution. For $i$ & $ii$ it did occur to me that the author was leading up to squaring/adding them but knowing to get it into that form in the first place has more to do with understanding what the end goal is and finding a way to twist it until it works. That comes only with experience. – Brian Vandenberg Sep 06 '21 at 05:16
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    Don't try to prevent extraneous solutions, filter them out at the end. – Trebor Sep 06 '21 at 05:55
  • @Trebor: We aren't solving anything; there are no solutions, extraneous or otherwise, in this problem. – Arturo Magidin Sep 06 '21 at 18:21

1 Answers1

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In this exercise, you are given a premise (the starting condition in the first sentence) and asked to prove a given consequence (the part of the sentence after "show that"). As such, as long as each step is valid, i.e., does not introduce a falsity, your proof is valid; and squaring both sides of an equation is always a valid step.

For example, in $$\sqrt x=2\iff x=4 \color{#C00}\implies x^2=16\iff x=\pm4,$$ the squaring step is logically valid. The issue is that in the context of solving equations, we need to ensure that each of our derived solutions indeed satisfies the equation that we started out with, i.e., that the given equation is a consequence of each of the derived solutions instead of merely vice versa. It is not the squaring per se, but the lack of “reversibility” in that step, that has created an extraneous solution. Other culprits include multiplying by $0,$ absolute-value terms, and implicit conditions on the given equation.

In practice, because trying to ensure “reversibility” at every step is tedious and prone to carelessness, when solving equations, it is best to just work in a forward direction like this $$\sqrt x=2\implies x=4 \implies x^2=16\implies x=\pm4,$$ then plugging the candidate solutions into the given equation to eliminate any extraneous ones. With experience, one develops a better awareness of when it is necessary to perform this filtering step, and when it isn't.

P.S. Peppering mathematical writing with the $\implies$ symbol, as you do, is not good practice: too many connective symbols decreases readability, but more importantly, it's not even the symbol $(\therefore)$ that you probably actually mean. I would omit every single $\implies$ from your above presentation.

ryang
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  • Yes squaring step is valid as long as both sides are equal but taking square root involves extraneous solutions – Lalit Tolani Sep 06 '21 at 07:09
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    @LalitTolani Don't you instead mean that taking square root potentially discards solutions? Yes, and only when one is taking square root carelessly: $\big(x^2=16\implies x=4\big)$ is wrong and discards a solution, but $\big(x^2=16\implies |x|=4\big)$ is correct and preserves both solutions (noting that $\sqrt{x^2}=|x|$). Of course, all this is tangential to the main discussion. – ryang Sep 06 '21 at 07:25
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    @Abu Safwan Md Farhan: It may help to note that when we have $P \implies Q,$ there is possible "information lost" in going from $P$ to $Q.$ In the context of squaring both sides of an equation, this loss of information is reflected in more possibilities for the variable (i.e. our knowledge of what values are possible for the variable will be less precise). As for when you know you're safe in squaring, if both sides are known to (always) be non-negative, then no extraneous solutions arise when squaring; also if both sides are known to (always) be non-positive. This is because (continued) – Dave L. Renfro Sep 06 '21 at 07:41
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    the function defined by $f(x)=x^2$ with domain $x \geq 0$ is bijective (and also bijective with domain $x \leq 0),$ which you can see geometrically from the fact that each of the two sides of the graph of the parabola $y=x^2$ satisfies the horizontal line test. – Dave L. Renfro Sep 06 '21 at 07:45
  • @RyanG Yes I meant that only – Lalit Tolani Sep 06 '21 at 08:10
  • Regarding "*except* when eqn A doesn't actually have a solution", singling out this exception is not necessary because the empty set is a subset of every set. – Dave L. Renfro Sep 06 '21 at 14:39
  • Thanks for the catch. Correction: Regarding @DaveL.Renfro's first point (thanks for the nice two-pronged elaboration): indeed, whenever $\text{}⟹\text{},;\text{}⊆\text{}.\quad$ (But a word of warning: when eqn A doesn't actually have a solution—as in the example in point 6 of this answer—eqn B may imply that $x$ is either $-1$ or $\frac13$ (i.e., at least one) when in fact $x$ is neither. Such anomalies, where every solution of eqn B is extraneous, happen due to deductive explosion.) – ryang Sep 06 '21 at 15:06
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    Multiplying both sides of an equation by $0$ can create a nuclear explosion (!), for example if each side of $e^x = 0$ is multiplied by $0.$ This is due to the highly non-injective property of the function $f(x) = 0 \cdot x.$ (I used to use the example of multiplying both sides of an equation by $0$ in certain classes I taught.) – Dave L. Renfro Sep 06 '21 at 15:13