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A hyperbola is a fascinating open curve formed by intersecting a plane with both halves of a double cone. Unlike ellipses or circles, hyperbolas consist of two separate branches that open either vertically or horizontally. The equation for a hyperbola centered at the origin can take two standard forms:

• Horizontal transverse axis: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
• Vertical transverse axis: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)

For the problem at hand, the equation is \( x^2 - y^2 = 12 \), indicating a horizontal transverse axis since there is no division affecting the coefficients of \( x^2 \) and \( y^2 \). This particular form will intersect at points along the x-axis further from the center and will open in opposite directions. Understanding this helps to grasp the nature of the points located on the curve, including the one provided in the problem.

Implicit differentiation is a technique used when dealing with equations that are not easily expressed as a function \( y = f(x) \). To differentiate implicitly, you treat \( y \) as a dependent variable and differentiate every term with respect to \( x \).

• Recall that \( \frac{d}{dx}(y) = \frac{dy}{dx} \).
• When performing differentiation, apply the derivative rules to both sides of the equation.

In the given problem, differentiating the equation \( x^2 - y^2 = 12 \) with respect to \( x \) gives \( 2x - 2y \frac{dy}{dx} = 0 \). The goal is to isolate \( \frac{dy}{dx} \), which represents the slope of the tangent line at any point \( (x, y) \) on the hyperbola. Successfully solving this helps us find how steep the curve is at specific points on the hyperbola.

Finding the equation of a tangent line involves determining the slope at a specific point on the curve, followed by using that slope in a linear equation format. A tangent line barely touches the curve at just one point, following the direction of that slope.

• First, find the slope \( m \) by evaluating \( \frac{dy}{dx} \) at the desired point.
• In this exercise, substituting \( x = 4 \) and \( y = 2 \) into \( \frac{dy}{dx} = \frac{x}{y} \) yields a slope of 2.

These calculations serve as the basis for establishing the tangent line's equation. Tangent lines are helpful as approximations of the curve near the point of tangency, making them vital in calculus and mathematical modeling.

The point-slope form is a straightforward way to write the equation of a line when you know a point on the line, \( (x_1, y_1) \), and the slope \( m \). The formula is:\[ y - y_1 = m(x - x_1) \]

• \( (x_1, y_1) \) is the given point on the line.
• \( m \) is the slope of the line.

In this exercise, with a point \((4, 2)\) and slope \(2\), the equation simplifies to: \[ y - 2 = 2(x - 4) \] This form is very intuitive for establishing a straight line equation and transitioning it to slope-intercept form \( y = mx + b \) after further simplification. Such translations are common practice in deriving and working with line equations, allowing for quick interpretation and usage.