Question

1. find the asymptotes of the hyperbola $\frac{(x - 1)^2}{4}-\frac{(y + 3)^2}{16}=1$

$y=-2x + 5$
$y=2x - 1$
$y=-2x - 5$
$y=2x - 5$
$y=-2x - 1$

Explanation:

Step1: Recall the standard form of hyperbola

For a hyperbola \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\), the asymptotes are given by \(y - k=\pm\frac{b}{a}(x - h)\). Here, \(h = 1,k=-3,a^2 = 4\) (so \(a = 2\)), \(b^2=16\) (so \(b = 4\)).

Step2: Calculate the asymptotes

Substitute the values into the formula.
For \(y - k=\frac{b}{a}(x - h)\):
\(y+3=\frac{4}{2}(x - 1)\)
\(y+3 = 2x-2\)
\(y=2x-5\)
For \(y - k=-\frac{b}{a}(x - h)\):
\(y + 3=-\frac{4}{2}(x - 1)\)
\(y+3=-2x + 2\)
\(y=-2x-1\)

Answer:

\(y = 2x-5\), \(y=-2x - 1\)