Questions: What is the vertex (are the vertices) of the shape formed by the equation (x-3)^2/49 - (y+5)^2/25 = 1?

Solution Steps

To find the vertices of the hyperbola given by the equation \(\frac{(x-3)^{2}}{49}-\frac{(y+5)^{2}}{25}=1\), we need to identify the center, the transverse axis, and the distance from the center to the vertices along the transverse axis.

1. Identify the center of the hyperbola, which is at \((h, k)\). For the given equation, \(h = 3\) and \(k = -5\).
2. Determine the distance \(a\) from the center to the vertices along the transverse axis. Here, \(a^2 = 49\), so \(a = 7\).
3. Since the hyperbola opens horizontally (because the \(x\)-term is positive), the vertices are located at \((h \pm a, k)\).

The given equation of the hyperbola is: \[ \frac{(x-3)^{2}}{49} - \frac{(y+5)^{2}}{25} = 1 \] The center \((h, k)\) of the hyperbola is \((3, -5)\).

The distance \(a\) from the center to the vertices along the transverse axis is found from the denominator of the \((x-h)^2\) term: \[ a^2 = 49 \implies a = 7 \]

Since the hyperbola opens horizontally, the vertices are located at: \[ (h \pm a, k) \] Substituting the values of \(h\), \(k\), and \(a\): \[ (3 + 7, -5) \quad \text{and} \quad (3 - 7, -5) \] Thus, the vertices are: \[ (10, -5) \quad \text{and} \quad (-4, -5) \]

Final Answer