Hyperbola
Standard form with translated axes
Grade 11
Question:
<p>The vertices of a hyperbola are at \((0, 0)\) and \((10, 0)\) and one of its foci is at \((18, 0)\). The equation of hyperbola is</p>
<p>(a) \(\frac{x^2}{25} - \frac{y^2}{144} = 1\)</p>
<p>(b) \(\frac{x^2}{25} - \frac{y^2}{144} = 1\)</p>
<p>(c) \(\frac{(x-5)^2}{25} - \frac{(y-5)^2}{144} = 1\)</p>
<p>(d) \(\frac{(x-5)^2}{25} - \frac{(y-5)^2}{144} = 1\)</p>
Step-by-Step Solution
Key Concept: Find the center as the midpoint of vertices, then use \(a\) and \(c\) to determine \(b\).
<p><strong>Solution:</strong> The vertices are at \((0, 0)\) and \((10, 0)\), so the center is at the midpoint \((5, 0)\).</p><p>The distance from center to vertex is \(a = 5\).</p><p>One focus is at \((18, 0)\), so the distance from center to focus is \(c = 18 - 5 = 13\).</p><p>Using \(b^2 = c^2 - a^2 = 169 - 25 = 144\).</p><p>Since the transverse axis is horizontal, the equation is \(\frac{(x-5)^2}{25} - \frac{(y-5)^2}{144} = 1\).</p><p>∴ Answer is (c).</p>
Correct Answer: C