Question

Let \(H\) be the right branch of the hyperbola \(x^{2}-y^{2}=1\) and let \(\ell\) be the line \(y=m(x-2)\) that passes through the point (2,0) with slope \(m,\) where \(-\inftyu(m)>1 .\) For what values of \(m\) are there two intersection points? b. Evaluate \(\lim _{m \rightarrow 1^{+}} u(m)\) and \(\lim _{m \rightarrow 1^{+}} v(m)\) c. Evaluate \(\lim _{m \rightarrow \infty} u(m)\) and \(\lim _{m \rightarrow \infty} v(m)\) d. Evaluate and interpret \(\lim _{m \rightarrow \infty} A(m)\)

Short answer

Answer: As the slope of the line 𝑙 approaches infinity, the area of the region R enclosed by the hyperbola H and the line 𝑙 in the first quadrant converges to a finite value. This means that as the line becomes more vertical, the area enclosed by the line and the hyperbola still has a specific value.

Step-by-step solution

Setting up Hyperbola and Line Equations

The equation of the right branch of the hyperbola is \(x^2 - y^2 = 1.\) The line \(\ell\) has the equation \(y=m(x-2).\)

Finding Intersection Points x-coordinates as functions of m

To find the intersection points, we need to solve the system of equations for \(x\) using the substitution method. Substitute \(y\) from the line equation into the hyperbola equation: \(x^2 - (m(x-2))^2 = 1.\)

Solving for x

Expanding the equation and simplifying, we get: \(x^2 - (m^2x^2 - 4mx^2 + 4m^2) = 1,\) which further simplifies to \((1-m^2)x^2 + 4mx^2 - 4m^2 = 1.\) Now, solve the equation for \(x^2:\) \((1-m^2 + 4m)x^2 = 1 + 4m^2,\) so \(x^2 = \frac{1+4m^2}{1-m^2+4m}.\) Since \(x > 0\) in the first quadrant, we get \(x = \pm \sqrt{\frac{1+4m^2}{1-m^2+4m}}.\)

Defining u(m) and v(m)

Let \(u(m) = -\sqrt{\frac{1+4m^2}{1-m^2+4m}}\) and \(v(m) = \sqrt{\frac{1+4m^2}{1-m^2+4m}}.\) The intersection points can only be in the first quadrant when \(1-m^2+4m > 0.\) Solving for \(m,\) we find that \(-\infty < m < 1.\)

Evaluating Limits

Now, we will evaluate the following limits: a. \(\lim_{m \rightarrow 1^{+}} u(m) = -1\) b. \(\lim_{m \rightarrow 1^{+}} v(m) = 1\) c. \(\lim_{m \rightarrow \infty} u(m) = 0\) d. \(\lim_{m \rightarrow \infty} v(m) = \infty\)

Evaluating the limit of A(m)

To evaluate the limit of the area function \(A(m),\) we need to set up an expression for the area of the region \(R.\) This area will be an integral of the line function minus the integral of the hyperbola: \(A(m) = \int_{u(m)}^{v(m)} (m(x-2) - \sqrt{x^2-1}) \, dx.\) Next, we evaluate the limit: \(\lim_{m \rightarrow \infty} A(m) = \lim_{m \rightarrow \infty} \int_{0}^{\infty} \sqrt{x^2-1} \, dx = \frac{1}{2}\int_{1}^{\infty} \frac{x^2-1}{x} \, dx\), which converges. Therefore, \(\lim_{m \rightarrow \infty} A(m)\) exists and is finite.

Interpretation

As the slope of the line \(\ell\) approaches infinity, the area of the region \(R\) between the hyperbola \(H\) and the line \(\ell\) converges to a finite value. This implies that as the line becomes more and more vertical, the area enclosed by the line and the hyperbola still has a specific value.

Key concepts

Hyperbola

A hyperbola is a type of conic section or a curve formed by the intersection of a cone with a plane. A real-world analogy of understanding a hyperbola can be seen in the way certain radio waves or satellite signals are sent and received. A hyperbola is characterized by its two branches, which open in opposite directions. The standard form of a hyperbola equation in this exercise is given by \(x^2 - y^2 = 1\). This particular equation describes a hyperbola oriented horizontally. The right branch consists of points where \(x^2\) values are greater than \(y^2\), meaning they meet the condition \(x^2 - y^2 = 1\).Understanding hyperbolas involves knowing:

• The center, which is the midpoint between the two vertices.
• The asymptotes, which are the lines that the curve approaches but never touches.

In the first quadrant, we only consider positive values of \(x\) and \(y\). This particular hyperbola only has an intersection in the first quadrant with certain lines.

Intersection of Curves

Finding where two curves intersect involves determining the points where their equations are equal. It's like finding the common ground between two different paths. In calculus, we work with their algebraic representation to determine these points precisely.For this exercise, the curves are represented by a hyperbola, \(x^2 - y^2 = 1\), and a line, \(y = m(x-2)\). To find the intersection, we substitute the line equation into the hyperbola, which leads to solving for \(x^2\) by expanding and rearranging: \[x^2 - (m(x-2))^2 = 1\]After simplification, we find the \(x\)-coordinates of the intersection points as functions of \(m\), named \(u(m)\) and \(v(m)\). It is essential that these solutions are constrained by the condition \(v(m) > u(m) > 1\), ensuring they are valid in our defined region.

Limits

In calculus, limits help us understand the behavior of functions as they approach specific points or infinity. They provide insight into trends and tendencies within complex systems.For this exercise, evaluating limits like \(\lim_{m \rightarrow 1^{+}} u(m) = -1\) and \(\lim_{m \rightarrow 1^{+}} v(m) = 1\), helps us anticipate the behavior of intersection points as the slope \(m\) approaches a particular value. Similarly, exploring \(\lim_{m \rightarrow \infty} u(m) = 0\) and \(\lim_{m \rightarrow \infty} v(m) = \infty\), shows how one intersection point approaches zero and the other moves towards infinity when \(m\) becomes infinitely large.These limits illustrate how the curves adjust and skew with changes in \(m\), impacting the nature and position of intersections.

Area Calculation

Calculating the area between curves in calculus uses integral calculus. It's like summing up all the minuscule slices of the region between the curves to find the total area.In the exercise, to find the area \(A(m)\) of region \(R\), one sets up the integral of the difference between the line and the hyperbola equations from one intersection point to the other:\[A(m) = \int_{u(m)}^{v(m)} (m(x-2) - \sqrt{x^2-1}) \, dx\]In essence, the area between a line and a curve in a specific region shows how widely they separate over that interval. By evaluating \(\lim_{m \rightarrow \infty} A(m)\), we determine that no matter how steep the line gets, the area will converge to a finite value. This indicates that the region does not grow indefinitely, despite the line's diverging path.