Problem 48 - Algebra and Trigonometry

Question

Graph the hyperbola whose equation is $$ 25 x^{2}-16 y^{2}-100 x-96 y-444=0 $$ Where are the foci located? What are the equations of the asymptotes?

Step-by-Step Solution

Answer

The foci of the hyperbola are located at (5.2,-3) and (-1.2,-3). The equations of the asymptotes are \(y = -3 + 1.25(x - 2)\) and \(y = -3 - 1.25(x - 2)\)

Step 1: Rewriting the Equation to Standard Form

The general form of a hyperbola is given as \(Ax^2 + By^2 + Dx + Ey + F = 0\). And the standard form is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) where center is (h,k), a is the length of the transverse axis and b is the length of the conjugate axis. Rewriting the given equation to resemble the standard form, you would get \(25(x-2)^2 - 16(y+3)^2 = 100\).

Step 2: Identify the Parameters

From the standard form, the center of the hyperbola can be found as (h,k) which is (2,-3) for this equation. You can also identify a and b, the lengths of the axes, where \(a^2 = \frac{100}{25} = 4\) and \(b^2 = \frac{100}{16} = 6.25\). This means a = 2 and b = 2.5, hence the length of transverse axis is 2 and the conjugate axis is 2.5 respectively.

Step 3: Determining the Foci

The distance from the center to each focus is given by \(c = \sqrt{a^2 + b^2}\) in the case of hyperbola. Using our values for a and b, we find \(c = \sqrt{4 + 6.25} = \sqrt{10.25} = 3.2\). Hence, the foci are at \((h+c, k)\) and \((h-c, k)\) which will be \((2+3.2, -3)\) and \((2-3.2, -3)\)

Step 4: Determine the Equation of Asymptotes

The equations of the asymptotes for a hyperbola are always in the form \(y = k \pm \frac{b}{a}(x - h)\). Substituting for h, k, a and b, we get \(y = -3 \pm \frac{2.5}{2}(x - 2)\). Hence the equations will be \(y = -3 + 1.25(x - 2)\) and \(y = -3 - 1.25(x - 2)\).