Question

1. using the equation \\( \frac { ( x - 1 ) ^ { 2 } } { 4 } - \frac { ( y + 3 ) ^ { 2 } } { 16 } = 1 \\), find \\( a, b, h \\) and \\( k \\).

\\( a = \\)
\\( b = \\)
\\( h = \\)
\\( k = \\)

Explanation:

Step1: Recall the standard form of a hyperbola

The standard form of a hyperbola \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1\) (opens left - right).

Step2: Compare the given equation with the standard form

Given equation \(\frac{(x - 1)^{2}}{4}-\frac{(y+3)^{2}}{16}=1\) can be written as \(\frac{(x - 1)^{2}}{2^{2}}-\frac{(y-(- 3))^{2}}{4^{2}}=1\).

• For \(a\):

Since \(a^{2}=4\), then \(a = 2\).

• For \(b\):

Since \(b^{2}=16\), then \(b = 4\).

• For \(h\):

Comparing \((x - h)\) with \((x - 1)\), we get \(h = 1\).

• For \(k\):

Comparing \((y - k)\) with \((y+3)=(y-(-3))\), we get \(k=-3\).

Answer:

\(a = 2\), \(b = 4\), \(h = 1\), \(k=-3\)