Question 629767
{{{(y - 2)^2/1^2 - (x- 4)^2/3^2 = 1}}}
The slopes of the asymptotes of a hyperbola are either <u>+</u>{{{a/b}}} or <u>+</u>{{{b/a}}}. If you're like me you have trouble which one to use when. This can be figured out, as you'll see shortly.<br>
From the equation we can tell that<ul><li>a = 1 (since {{{a^2}}} is always under the first term)</li><li>b = 3 (since {{{b^2}}} is always under the first term)</li><li>the hyperbola is vertically oriented (since the the first term has "y", not "x", in it).</li></ul>On a vertically oriented hyperbola, the a will be a vertical distance (IOW a "rise") and the b will be a horizontal distance (IOW, a "run"). Since slope is always rise/run we will need to use <u>+</u>{{{a/b}}} for this hyperbola.<br>
So the slopes of this particular hyperbola are <u>+</u>{{{1/3}}}