Question

Find the equations of the tangent and the normal lines to the given parabola at the given point. Sketch the parabola, the tangent line, and the normal line. $$ y^{2}=-15 x,(-3,-3 \sqrt{5}) $$

Short answer

Tangent: \( y = \frac{\sqrt{5}}{2}x - \frac{3\sqrt{5}}{2} \); Normal: \( y = -\frac{2}{\sqrt{5}}x - \frac{21}{\sqrt{5}} \).

Step-by-step solution

Understand the given equation

The given parabola is described by the equation \( y^2 = -15x \), which is a sideways parabola that opens to the left. The point provided is \((-3, -3\sqrt{5})\). Our task is to find the tangent and normal lines at this point on the parabola.

Find the derivative dy/dx

To find the slope of the tangent line to the parabola at the given point, we first need to find the derivative of \( y \) with respect to \( x \). Differentiating the equation implicitly gives us \( 2y \frac{dy}{dx} = -15 \), which simplifies to \( \frac{dy}{dx} = \frac{-15}{2y} \).

Calculate the slope at the given point

Substitute \( y = -3\sqrt{5}\) into the derivative formula: \( \frac{dy}{dx} = \frac{-15}{2(-3\sqrt{5})} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2} \). This is the slope of the tangent line at the point \((-3, -3\sqrt{5})\).

Write the equation of the tangent line

The equation of a line can be written as \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the point on the line and \( m \) is the slope. Substituting the values, we get: \( y + 3\sqrt{5} = \frac{\sqrt{5}}{2} (x + 3) \), which simplifies to \( y = \frac{\sqrt{5}}{2}x + \frac{3\sqrt{5}}{2} - 3\sqrt{5} \), or \( y = \frac{\sqrt{5}}{2}x - \frac{3\sqrt{5}}{2} \).

Find the slope of the normal line

The slope of the normal line is the negative reciprocal of the tangent line's slope. Thus, it is \( -\frac{2}{\sqrt{5}} \).

Write the equation of the normal line

Using the point-slope form again for the normal line, we have: \( y + 3\sqrt{5} = -\frac{2}{\sqrt{5}}(x + 3) \), which simplifies to \( y = -\frac{2}{\sqrt{5}}x - \frac{6}{\sqrt{5}} - 3\sqrt{5} \), or \( y = -\frac{2}{\sqrt{5}}x - \frac{21}{\sqrt{5}} \).

Sketch the curves

Draw the sideways-opening parabola \( y^2 = -15x \). Then, draw the tangent line \( y = \frac{\sqrt{5}}{2}x - \frac{3\sqrt{5}}{2} \) and the normal line \( y = -\frac{2}{\sqrt{5}}x - \frac{21}{\sqrt{5}} \) at the point \((-3, -3\sqrt{5})\) on the parabola.

Key concepts

Parabola

A parabola is a significant curve in mathematics that is defined as a set of points equidistant from a fixed point called the "focus" and a line called the "directrix." In this exercise, the parabola is given by the equation \( y^2 = -15x \), which is unique because it opens sideways to the left. This happens when the square term is on the \( y \) variable rather than \( x \). To understand the shape better, consider plotting several points that satisfy the equation to ascertain the curve's orientation and trajectory.
Usually, parabolas that open left or right are involved when the equation has the form \( y^2 = kx \), where \( k \) decides the direction and width of the opening. In our specific equation, \( k = -15 \), indicating the parabola opens to the left.

Derivative

The derivative in calculus represents the rate at which a function is changing at any point. For curves, this manifests as the slope of the tangent line at a particular point. To find this slope for the given parabola at a specific point, we need to differentiate the equation. Here, implicit differentiation is used to compute \( \frac{dy}{dx} \). Implicit differentiation is essential when dealing with equations where \( y \) is not isolated.
Executing this differentiation on our parabola, \( y^2 = -15x \), we apply the derivative to both sides concerning \( x \), treating \( y \) as a function of \( x \). This technique is beneficial because it allows us to find \( \frac{dy}{dx} = \frac{-15}{2y} \). From here, plugging in the specific \( y \) value at the given point will yield the slope of the tangent.

Slope of Tangent Line

The slope of the tangent line expresses how steeply the line rises or falls at the given point on the curve. For the parabola, the tangent line signifies the line that touches the curve precisely at one point and shares the curve's initial direction at that point.
To find this, we calculate the derivative \( \frac{dy}{dx} \) and substitute the \( y \) value of the given point to derive the actual slope. This exercise provided the point \((-3, -3\sqrt{5})\), where substituting \( y = -3\sqrt{5} \) resulted in the slope \( \frac{\sqrt{5}}{2} \). This value provides the necessary slope to formulate the equation of the tangent line using the point-slope formula:

• Equation: \( y - y_1 = m(x - x_1) \)
• Where \( m \) is the slope, and \((x_1, y_1)\) is the point of tangency.

Implicit Differentiation

Implicit differentiation is a vital tool in calculus used for equations that do not easily solve for \( y \) explicitly in terms of \( x \). Our scenario is perfect for implicit differentiation because of the form \( y^2 = -15x \). Instead of isolating \( y \, (\)which can become complex,\()\) we differentiate both sides of the equation with respect to \( x \) directly.
This process involves taking the derivative of each term, recalling that \( y \) is treated as a function of \( x \), and applying the chain rule where necessary. For \( 2y \cdot \frac{dy}{dx} \) from \( y^2 \), differentiate directly leading to \( \frac{dy}{dx} = \frac{-15}{2y} \). This straightforward calculation is powerful in finding slopes and understanding behaviors of growth within curves that are implicitly defined, and crucial for achieving exact results, such as tangent and normal lines of curves in complex orientations.