91Ó°ÊÓ

A hyperbola is a type of conic section, similar to ellipses and parabolas. They appear as two symmetrical open curves on a plane. In general, hyperbolas are defined as the set of points where the difference of the distances to two fixed points, called foci, is constant. The standard form of a hyperbola can be written as \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1,\]where

• \( h \) and \( k \) are the coordinates of the center of the hyperbola.
• \( a \) and \( b \) are the distances from the center to the vertices along the transverse and conjugate axes, respectively.

In this exercise, the equation for the hyperbola was given as \[ 4x^2 - 9y^2 - 8x + 6y - 36 = 0.\]By completing the square for both \( x \) and \( y \), we can transform the equation into a form closer to the standard hyperbola equation. Understanding hyperbolas helps in visualizing the curves and their geometric properties, which are crucial for finding tangent and normal lines.

Implicit differentiation is a technique used when dealing with equations where the dependent variable is not isolated. For functions that can be written as \( y = f(x) \), finding the derivative can be straightforward. However, in cases where \( y \) and \( x \) are mixed together, we use implicit differentiation.To perform implicit differentiation, you differentiate each term of the equation with respect to \( x \), remembering to multiply by \( \frac{dy}{dx} \) when differentiating terms involving \( y \). This is because \( y \) is considered a function of \( x \).In our problem, the derivative of the equation \( 4x^2 - 9y^2 - 8x + 6y - 36 = 0 \) with respect to \( x \) is found by differentiating every term:

• \( 8x \) comes from \( 4x^2 \).
• \(-18y\frac{dy}{dx}\) from \(-9y^2\).
• \(-8\) from \(-8x\).
• \(6\frac{dy}{dx}\) from \(6y\).

After simplifying, we solve for \( \frac{dy}{dx} \). This derivative gives us the slope of the tangent line at the point of interest.

The point-slope form of a line's equation is incredibly useful when you have the slope of the line and a point on the line. It can be expressed as\[ y - y_1 = m(x - x_1), \]where:

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

Once the slope \( m \) is determined, you can directly substitute the values into the formula. In this exercise, the hyperbola's implicit differentiation provided the slope at point \((-3, 2)\), which is \(-\frac{16}{15}\) for the tangent line.Using the point-slope form:- For the tangent line: \[ y - 2 = -\frac{16}{15}(x + 3). \]- For the normal line: find the negative reciprocal of the tangent slope, which is \( \frac{15}{16} \). This changes the tangent's direction to be perpendicular, allowing us to write: \[ y - 2 = \frac{15}{16}(x + 3). \]The point-slope form makes it easy to derive the final equation of the line in other standard forms, depending on what's needed, expanding its versatility in solving geometric problems.