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If there is a point (a,b) on the hyperbola $x^2-y^2=4$ that is closest to the point (0,1), then what is this point?

I prefer this problem to be solved with the knowledge of the the maximum and minimum using the derivatives and so on.

I am a bit confused

I said that I will take the square distance between the point $0,1$ and $y=$$\sqrt{x^2-4}$

then I got $D(x)=x^2-(\sqrt{x^2-4}-1)^2$

after that I took the derivative and made it equal to zero and I was lost at this point since I didn't know how to get the values that would lead me to the Minimum.

?!?!

Thomas Andrews
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Maher
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  • There is definitely a closest point, because the hyperbola is closed. – Thomas Andrews Dec 18 '14 at 21:29
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    Shouldn't $D(x)=x^2+(\sqrt{x^2-4}-1)^2$? The minus is only in the hyperbola definition, not in the definition of Euclidean distance. Of course, $y=\pm\sqrt{x^2-4}$, but the positive $y$ is obviously closer. – Thomas Andrews Dec 18 '14 at 21:31

3 Answers3

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It's probably easier to take $x=\pm\sqrt{y^2+4}$, since this is true for all $y$, while looking at the question in terms of $x$ requires the condition that $|x|\geq 4$.

Also, $x^2+(y-1)^2 = y^2+4 +(y-1)^2$ is much easier to minimize, and doesn't depend on the sign of $x$.

Thomas Andrews
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  • Great. But I still want to know what is wrong with my previous solution? – Maher Dec 18 '14 at 21:44
  • You didn't give a complete solution, so it's hard to tell what was wrong with it. As I mentioned in the comment above, you got a sign wrong in $D(x)$. The real problem with your solution is that, by choosing the wrong variable to replace, you gave yourself a much harder algebraic problem. Without you writing it out, it's hard to see where your error was, aside from the sign error (which you did write out.) – Thomas Andrews Dec 18 '14 at 21:46
  • should I include +and- inside the squared parentheses? – Maher Dec 18 '14 at 21:49
  • You've given zero context for that last question, @Maher. Write it out. – Thomas Andrews Dec 18 '14 at 21:49
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here is one way to compute the closes distance. let the foot of the perpendicular from $(0,1)$ to $x^2-y^2 = 4$ be $(a,b)$ the slope of the line is $b - 1 \over a$ and the slope of hyperbola is $a \over b,$ so $a, b$ satisfy:${(b-1) \over a}{ a \over b} = -1$ this gives $$a = 0, b = { 1 \over 2}.$$ $a = 0$ is not in the domain so we have $b = {1 \over 2} \ and \ a = \sqrt{17}/2.$

so the shortest distance from $(0,1)$ to the $x^2 - y^2 = 4$ is $$\sqrt{(1-1/2)^2 + (0 \pm\sqrt{17/2})^2 } = {3\sqrt 2 \over 2}. $$

abel
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WLOG, we can represent any point on the given hyperbola as $x=2\sec A,y=2\tan A$

So, we need to minimize $\sqrt{(2\sec A-0)^2+(2\tan A-1)^2}=\sqrt{8\tan^2A-4\tan A+5}$

Now, $8\tan^2A-4\tan A+5=\dfrac{16\tan^2A-8\tan A+10}2=\dfrac{(4\tan A-1)^2+9}2\ge\dfrac92$

lab bhattacharjee
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