Determine the coefficients a, b, c, and d such that

Determine the coefficients $a, b, c,$ and $d$ such that the hyperbola $y = \frac{ax + b}{cx + d}$ passes through the origin and has asymptotes $x = 2$ and $y = 1$. Then calculate the area bounded by the arc of the resulting hyperbola and the line $x + y + 1 = 0$.

We first determine the hyperbola's equation, then compute the enclosed area.

Part 1: Find the coefficients

Step 1: Use the condition that the hyperbola passes through the origin

At $(0, 0)$:

\[0 = \frac{a \cdot 0 + b}{c \cdot 0 + d} = \frac{b}{d}\]

Therefore: $b = 0$. So the hyperbola is: $y = \frac{ax}{cx + d}$

Step 2: Use the vertical asymptote $x = 2$

The vertical asymptote occurs where the denominator is zero:

\[cx + d = 0 \implies x = -\frac{d}{c}\]

Given that $x = 2$:

\[-\frac{d}{c} = 2 \implies d = -2c\]

Step 3: Use the horizontal asymptote $y = 1$

For large $|x|$, the horizontal asymptote is:

\[y = \lim_{x \to \infty} \frac{ax}{cx + d} = \lim_{x \to \infty} \frac{ax}{cx(1 + d/(cx))} = \frac{a}{c}\]

Given that $y = 1$:

\[\frac{a}{c} = 1 \implies a = c\]

Step 4: Express the hyperbola

Using $a = c$ and $d = -2c$:

\[y = \frac{cx}{cx - 2c} = \frac{cx}{c(x - 2)} = \frac{x}{x - 2}\]

We can choose $c = 1$ (scaling factor), giving:

\[\boxed{a = 1, \quad b = 0, \quad c = 1, \quad d = -2}\]

The hyperbola is:

\[\boxed{y = \frac{x}{x - 2}}\]

Part 2: Calculate the area

Step 1: Find intersection points with the line $x + y + 1 = 0$

From the line: $y = -x - 1$. Setting equal to the hyperbola:

\[\frac{x}{x - 2} = -x - 1\]

\[x = (-x - 1)(x - 2)\]

\[x = -x^2 + 2x - x + 2\]

\[x = -x^2 + x + 2\]

\[x^2 = 2\]

\[x = \pm\sqrt{2}\]

When $x = \sqrt{2}$: $y = -\sqrt{2} - 1$. Point: $(\sqrt{2}, -\sqrt{2} - 1)$

When $x = -\sqrt{2}$: $y = \sqrt{2} - 1$. Point: $(-\sqrt{2}, \sqrt{2} - 1)$

Step 2: Determine which curve is above

The hyperbola has a vertical asymptote at $x = 2$. For $x < 2$, let's check at $x = 0$:

Hyperbola: $y = \frac{0}{0 - 2} = 0$. Line: $y = -0 - 1 = -1$

So the hyperbola is above the line for $x$ near 0. For $x \in [-\sqrt{2}, \sqrt{2}]$ (both less than 2), the hyperbola is above the line.

Step 3: Set up the integral

\[A = \int_{-\sqrt{2}}^{\sqrt{2}} \left[\frac{x}{x - 2} - (-x - 1)\right] dx\]

\[= \int_{-\sqrt{2}}^{\sqrt{2}} \left[\frac{x}{x - 2} + x + 1\right] dx\]

Step 4: Simplify the integrand

\[\frac{x}{x - 2} + x + 1 = \frac{x + (x + 1)(x - 2)}{x - 2}\]

\[= \frac{x + x^2 - 2x + x - 2}{x - 2} = \frac{x^2 - 2}{x - 2}\]

Step 5: Simplify further by polynomial division

Writing $x^2 - 2 = x(x - 2) + 2(x - 1)$, then $2(x-1) = 2(x-2) + 2$:

\[\frac{x^2 - 2}{x - 2} = x + \frac{2(x - 1)}{x - 2} = x + \frac{2(x-2) + 2}{x-2} = x + 2 + \frac{2}{x - 2}\]

Step 6: Integrate

\[A = \int_{-\sqrt{2}}^{\sqrt{2}} \left[x + 2 + \frac{2}{x - 2}\right] dx\]

\[= \left[\frac{x^2}{2} + 2x + 2\ln|x - 2|\right]_{-\sqrt{2}}^{\sqrt{2}}\]

At $x = \sqrt{2}$:

\[\frac{2}{2} + 2\sqrt{2} + 2\ln|\sqrt{2} - 2| = 1 + 2\sqrt{2} + 2\ln(2 - \sqrt{2})\]

At $x = -\sqrt{2}$:

\[\frac{2}{2} - 2\sqrt{2} + 2\ln|-\sqrt{2} - 2| = 1 - 2\sqrt{2} + 2\ln(2 + \sqrt{2})\]

Therefore:

\[A = [1 + 2\sqrt{2} + 2\ln(2 - \sqrt{2})] - [1 - 2\sqrt{2} + 2\ln(2 + \sqrt{2})]\]

\[= 4\sqrt{2} + 2\ln(2 - \sqrt{2}) - 2\ln(2 + \sqrt{2})\]

\[= 4\sqrt{2} + 2\ln\frac{2 - \sqrt{2}}{2 + \sqrt{2}}\]

\[\boxed{A = 4\sqrt{2} + 2\ln\frac{2 - \sqrt{2}}{2 + \sqrt{2}} \text{ square units}}\]

Or approximately: $A \approx 5.657 - 1.763 \approx 3.894$ square units

Summary

Coefficients:

\[\boxed{a = 1, \quad b = 0, \quad c = 1, \quad d = -2}\]

Hyperbola:

\[\boxed{y = \frac{x}{x - 2}}\]

Area:

\[\boxed{A = 4\sqrt{2} + 2\ln\frac{2 - \sqrt{2}}{2 + \sqrt{2}} \approx 3.89 \text{ square units}}\]

Key steps:

1. Use the origin condition to get $b = 0$
2. Use the vertical asymptote to get $d = -2c$
3. Use the horizontal asymptote to get $a = c$
4. Normalize $c = 1$ to get $y = \frac{x}{x-2}$
5. Find intersections with the line at $x = \pm\sqrt{2}$
6. Simplify the rational integrand by division: $\frac{x^2-2}{x-2} = x + 2 + \frac{2}{x-2}$
7. Integrate to get $A = 4\sqrt{2} + 2\ln\frac{2-\sqrt{2}}{2+\sqrt{2}}$

Key insight:

Three geometric conditions (a point and two asymptotes) are exactly enough to fix a Möbius-type rational function up to an overall scale factor.