Let T and C respectively be the transverse and conjugate axes of the hyperbola 16x² - y² + 64x + 4y

Q: Wait, why is the transverse axis $y=2$ and not $x=-2$?

Great question! In the standard form of a hyperbola, $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$, the transverse axis is the line that passes through the vertices and the center. Since our $x$ term is positive, the hyperbola opens horizontally, meaning the transverse axis is the horizontal line $y=k$. In our case, $k=2$, so the transverse axis is $y=2$.

Q: How do I know which curve is the upper boundary for the integral?

Think about the region constraints. We are looking for the area *below* the transverse axis $T$ ($y=2$) and *above* the parabola ($y=x^2-4$). Since the region is trapped between these two, the line $y=2$ must be the upper boundary ($y_{upper}$) and the parabola $y=x^2-4$ must be the lower boundary ($y_{lower}$).

Q: Why did we choose $x = \sqrt{6}$ and not $-\sqrt{6}$ as the right limit?

The problem explicitly states that the region is to the *right* of the conjugate axis $C$. The conjugate axis is the vertical line $x=-2$. Therefore, we only consider the portion of the region where $x > -2$. Since the intersection points of $y=2$ and $y=x^2-4$ are $x = \pm\sqrt{6}$, we must reject $-\sqrt{6}$ because it is less than $-2$.

Question

Let T and C respectively be the transverse and conjugate axes of the hyperbola $16x^2 - y^2 + 64x + 4y + 44 = 0$. Then the area of the region above the parabola $x^2 = y + 4$, below the transverse axis T and on the right of the conjugate axis C is:

Options:

(A) $4\sqrt{6} + \frac{44}{3}$

(B) $4\sqrt{6} + \frac{28}{3}$

(C) $4\sqrt{6} - \frac{44}{3}$

(D) $4\sqrt{6} - \frac{28}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering the hidden geometry of a hyperbola and a parabola.

It is easy to look at an equation like $16x^2 - y^2 + 64x + 4y + 44 = 0$ and feel overwhelmed. But remember, every complex equation is just a story waiting to be told. Let us break it down, step by step, and find the beauty in the calculation.

### Taming the Hyperbola

Our first task is to bring order to chaos. The equation $16x^2 - y^2 + 64x + 4y + 44 = 0$ is in a general form that hides the hyperbola's true nature. To reveal its center and axes, we must complete the square.

We group the $x$ terms and the $y$ terms separately:

$$16(x^2 + 4x) - (y^2 - 4y) + 44 = 0$$

By completing the square for $x$, we add and subtract $4$ inside the bracket, transforming $x^2 + 4x$ into $(x+2)^2 - 4$. Similarly, for $y$, we transform $y^2 - 4y$ into $(y-2)^2 - 4$.

When we distribute the constants and simplify, we arrive at the elegant standard form:

$$\frac{(x+2)^2}{1} - \frac{(y-2)^2}{16} = 1$$

This is the moment of clarity. We can now see that the center of our hyperbola is at $(-2, 2)$. The transverse axis $T$, which is the horizontal line passing through the center, is $y=2$. The conjugate axis $C$, the vertical line passing through the center, is $x=-2$.

### Visualizing the Region

Now, let us introduce the parabola $x^2 = y + 4$, or more simply, $y = x^2 - 4$. This is an upward-opening parabola with its vertex at $(0, -4)$.

The problem asks us to find the area of a specific region: above this parabola, below the transverse axis $T$ ($y=2$), and to the right of the conjugate axis $C$ ($x=-2$).

Imagine standing at the center of the hyperbola $(-2, 2)$. You are looking at a region bounded on the left by the vertical line $x=-2$, on the top by the horizontal line $y=2$, and on the bottom by the curve $y=x^2-4$.

The region ends on the right where the parabola meets the line $y=2$. Setting $x^2 - 4 = 2$, we find $x^2 = 6$, which gives us $x = \sqrt{6}$ (since we are restricted to the right of $x=-2$).

### The Calculus of Area

We have our boundaries: $x$ ranges from $-2$ to $\sqrt{6}$. The area $A$ is the integral of the upper curve minus the lower curve:

$$A = \int_{-2}^{\sqrt{6}} (y_{upper} - y_{lower}) dx$$

Substituting our boundaries, we get:

$$A = \int_{-2}^{\sqrt{6}} (2 - (x^2 - 4)) dx = \int_{-2}^{\sqrt{6}} (6 - x^2) dx$$

This integral is the heart of our problem. It represents the accumulation of vertical strips from the left boundary to the right boundary. Let us perform the integration:

$$\left[ 6x - \frac{x^3}{3} \right]_{-2}^{\sqrt{6}}$$

### Final Calculation

Now, we evaluate the definite integral. First, the upper limit $\sqrt{6}$:

$$6\sqrt{6} - \frac{(\sqrt{6})^3}{3} = 6\sqrt{6} - \frac{6\sqrt{6}}{3} = 6\sqrt{6} - 2\sqrt{6} = 4\sqrt{6}$$

Next, the lower limit $-2$:

$$6(-2) - \frac{(-2)^3}{3} = -12 - \frac{-8}{3} = -12 + \frac{8}{3} = -\frac{28}{3}$$

Subtracting the lower limit from the upper limit gives us the final area:

$$A = 4\sqrt{6} - \left( -\frac{28}{3} \right) = 4\sqrt{6} + \frac{28}{3}$$

And there it is! The area of the region is **$4\sqrt{6} + \frac{28}{3}$**. It is a beautiful result, born from the marriage of algebra and calculus.

Select Answer:

Visualized Solution

Analyze the Hyperbola Equation

Completing the Square for $x$

Completing the Square for $y$

Standard Form of Hyperbola

Identifying Axes $T$ and $C$

The Parabola Boundary

Finding Intersection Points

Setting up the Integral

Integration and Evaluation

Final Answer Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Taming the Hyperbola

Visualizing the Region

The Calculus of Area

Final Calculation

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Select Answer:

Visualized Solution

Analyze the Hyperbola Equation

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