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Question:

Consider the following part of a hyperbola:

$$x^2 - y^2 = 1, ~~~~~~~x\geq0$$

What is the shortest distance between a point on the hyperbola and the point P = (0,1)?

Attempted answer:

The distance between a point on the hyperbola and P for this scenario is:

$$D^2 = x^2 + (y-1)^2$$

Taking the hyperbola equation and solving for $x^2$:

$$x^2 - y^2 = 1 \Rightarrow x^2 = 1 + y^2$$

Putting this into the equation for $D^2$:

$$D^2 = 1 + y^2 + y^2 - 2y + 1 = 2y^2-2y +2$$

Taking the derivative (we can here minimize $D^2$ for simplicity):

$$\frac{d D^{2}}{dx} = 4y - 2$$

Setting the derivative to zero and solving for y:

$$\frac{d D^{2}}{dx} = 0 \Rightarrow 4y - 2 = 0 \Rightarrow y = \frac{1}{2}$$

This has to be a minimum because the distance in a hyperbola of course would go towards plus or minus infinity.

Finding x from the hyperbola equation:

$$x = \sqrt{1+y^2} = \sqrt{1+\frac{1}{4}} = \sqrt{\frac{5}{4}}$$

Finding the distance:

$$D = \sqrt{\frac{5}{4}}^2 + (\frac{1}{2} - 1)^2 = \frac{6}{4}$$

However, the expected answer is $\sqrt{\frac{3}{2}}$. What has gone wrong?

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  • $\begingroup$ Use tangent and normal concept $\endgroup$ Jul 24 '19 at 11:02
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You did it right but at last what you obtain is $D^2$ not $D$

So you got $$D^2=\frac54 + \frac14 =\frac64 =\frac32$$ Thus, $$D=\sqrt{\frac32}$$

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We can avoid calculus altogether

$$2y^2-2y+2=2\left(y-\dfrac12\right)^2+2-2\left(\dfrac12\right)^2\ge2-2\left(\dfrac12\right)^2$$

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