Question

find the standard form equation for the hyperbola centered at the origin that has vertices $(\pm4,0)$ and foci $(\pm6,0)$. standard form

Explanation:

Step1: Determine the form of the hyperbola

Since the vertices and foci are on the \(x -\)axis (\((\pm a,0)\) and \((\pm c,0)\)), the standard form of the hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1\).
Given \(a = 4\) (from vertices \((\pm4,0)\)) and \(c = 6\) (from foci \((\pm6,0)\)).

Step2: Find \(b^{2}\) using the relationship \(c^{2}=a^{2}+b^{2}\)

Substitute \(a = 4\) and \(c = 6\) into the formula \(c^{2}=a^{2}+b^{2}\).

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Step3: Write the standard - form equation

Substitute \(a^{2}=16\) and \(b^{2}=20\) into \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\).
The equation is \(\frac{x^{2}}{16}-\frac{y^{2}}{20}=1\)

Answer:

\(\frac{x^{2}}{16}-\frac{y^{2}}{20}=1\)