Calculating Designs

When y=2

$$4x^2=1+4/9$$

$${\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}} = {\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{4}}}{{\mathtt{9}}}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{13}}}}}{{\mathtt{6}}}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\mathtt{13}}}}}{{\mathtt{6}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.600\: \!925\: \!212\: \!577\: \!331\: \!5}}\\
{\mathtt{x}} = {\mathtt{0.600\: \!925\: \!212\: \!577\: \!331\: \!5}}\\
\end{array} \right\}$$

width at the top =  $${\frac{{\sqrt{{\mathtt{13}}}}}{{\mathtt{3}}}}$$    metres

It appears to me that we all got different answers.  

I for one did not really understand the question.  

Now Chris's is the same as mine (he did a little correction) but Heueka's is still different.

Maybe Heurka  interpreted it differently from me and Chris?

An architect's design for a building includes some large pillars with cross sections in the shape of hyperbolas. The curves are modeled by . If the pillars are 4 meters tall, find the width of the top of each pillar.

$$\dfrac{x^2}{0.25} - \dfrac{y^2}{9} = 1 \quad \text{we need x}\\\\
\dfrac{x^2}{0.25} = 1 + \dfrac{y^2}{9} \\\\
x^2 = 0.25*(1 + \dfrac{y^2}{9}) \quad | \quad \pm\sqrt{} \\\\
x_{1,2} = \pm 0.5 * \sqrt{ (1 + \dfrac{y^2}{9} ) } \\ \\
x_{1,2} = \pm \dfrac{ 0.5}{3} * \sqrt{ y^2 + 9} \\ \\
x_{1,2} = \pm \dfrac{ 1}{6} * \sqrt{ y^2 + 9} \\ \\
x_1= -\dfrac{ 1}{6} * \sqrt{ y^2 + 9} \\\\
x_2= \dfrac{ 1}{6} * \sqrt{ y^2 + 9}$$

$$width = x_2 - x_1 = \dfrac{ 1}{6} * \sqrt{ y^2 + 9 }- \left(-\dfrac{1}{6} * \sqrt{ y^2 + 9 } \right) \\\\
width = 2*\dfrac{1}{6} * \sqrt{ y^2 + 9 } \\\\
width = \dfrac{1}{3} * \sqrt{ y^2 + 9 } \quad | \quad y = 4\ meter\\\\
width = \dfrac{1}{3} * \sqrt{ 16 + 9 } \\\\
width = \dfrac{1}{3} * \sqrt{ 25 } \\\\
width = \dfrac{1}{3} * 5 \\\\
width = \dfrac{5}{3} \\\\
width = 1.\overline{6} \ m$$

We have

(x^2 / .25) - (y^2 / 9 )  = 1

And, let's assume that the base of the pillar lies on the line y = -2

Then, since the pillars are 4 m high, at the top of the pillar, y = 2

So we have

x^2 / .25  - 2^9 / 9 = 1

x^2 / .25 - 4/9  = 1        add 4/9 to each side

x^2 / .25  = 1 + 4/9

x^2 / .25  = 13 / 9        multiply by .25 on both sides

x^2 = (.25 * 13) / 9

x^2 = 13/36

x = ±√(13) / 6

And these are the two x values on the hyperbola when y = 2

So.....the distance between these two x values is the width of the cross-section at the top of the pillar =  (2)√(13)/6 = (1/3)√(13) =  aboiut 1.2019 m

Here's a graph of the situation.........https://www.desmos.com/calculator/ts3acz4ffr

Sorry, Melody......I mucked it up originally by reading ".25" as "25"..............you understood it as I did....

Melody....I think the difference in our answers lies in the fact that you and I have assumed that the base of the pillar lies on the line y = -2  while heureka has assumed that it lies on the x axis......either orientation could be correct...