Problem 63 - Algebra and Trigonometry

Question

An architect designs two houses that are shaped and positioned like a part of the branches of the hyperbola whose equation is \(625 y^{2}-400 x^{2}=250,000,\) where \(x\) and \(y\) are in yards. How far apart are the houses at their closest point?

Step-by-Step Solution

Answer

The shortest distance between the two designed houses is 8 yards.

Step 1: Rewriting the Hyperbola's Equation

We must first rewrite the given equation of the hyperbola in standard form. The standard form of hyperbola is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Rewriting the equation in standard form will help identify the values of \(a\) and \(b\).\n\nSo, we can rewrite \(625 y^2 - 400 x^2 = 250,000\) as\n \(\frac{y^2}{4^2} - \frac{x^2}{5^2} = 1\) by dividing whole equation by \(250,000\) which is nothing but \(a^2 b^2\).

Step 2: Identifying the Vertices

The vertices of a vertical hyperbola given in standard form are \((0, ± a )\). Here in our equation, \(a = 4\), so our vertices are \((0,4)\) and \((0,-4)\). These are the points at which the distance between two houses will be shortest.

Step 3: Calculating the Shortest Distance

Now we have the vertices, the minimum distance or shortest distance between the two houses will be the difference in y-coordinates of these vertices.\n\nSo, the shortest distance = 4 - (-4) = 8 yards.