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. $$ y^{2}=-9 x,(-1,-3) $$

Short answer

Tangent line: \( y = \frac{3}{2}x - \frac{3}{2} \), Normal line: \( y = -\frac{2}{3}x - \frac{11}{3} \).

Step-by-step solution

Understanding the Parabola Equation

The given parabola equation is \( y^2 = -9x \). This type represents a parabola that opens to the left because the coefficient of \( x \) is negative and \( y \) is squared.

Finding the Slope of the Tangent Line

To find the slope of the tangent line, we first differentiate the given equation implicitly with respect to \( x \). Differentiating, we have \( 2y \frac{dy}{dx} = -9 \). At the point \((-1, -3)\), substitute \( y = -3 \) to find \( \frac{dy}{dx} \): \[ 2(-3) \cdot \frac{dy}{dx} = -9 \Rightarrow -6 \cdot \frac{dy}{dx} = -9 \Rightarrow \frac{dy}{dx} = \frac{3}{2} \]. Hence, the slope of the tangent line is \( \frac{3}{2} \).

Equation of the Tangent Line

Using the point-slope form of the equation for a line, the tangent line at \((-1, -3)\) is given by \( y - (-3) = \frac{3}{2} (x - (-1)) \). Simplifying, the equation becomes \( y + 3 = \frac{3}{2}(x + 1) \). Distributing \( \frac{3}{2} \), we have \( y + 3 = \frac{3}{2}x + \frac{3}{2} \). Further simplifying gives \( y = \frac{3}{2}x - \frac{3}{2} \).

Finding the Slope of the Normal Line

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent line's slope. Thus, the slope of the normal line is \(-\frac{2}{3}\).

Equation of the Normal Line

Using the slope-point form for the normal line at \((-1, -3)\), we have \( y + 3 = -\frac{2}{3}(x + 1) \). Simplifying, the equation becomes \( y + 3 = -\frac{2}{3}x - \frac{2}{3} \). Further simplifying gives \( y = -\frac{2}{3}x - \frac{2}{3} - 3 \). Calculate \(-\frac{2}{3} - 3 = -\frac{11}{3} \), so the equation is \( y = -\frac{2}{3}x - \frac{11}{3} \).

Sketching the Graph

Draw the parabola \( y^2 = -9x \), which opens to the left with vertex at the origin. Mark the point \((-1, -3)\) on the parabola. Draw the tangent line \( y = \frac{3}{2}x - \frac{3}{2} \) through \((-1, -3)\) with a slope of \( \frac{3}{2} \). Draw the normal line \( y = -\frac{2}{3}x - \frac{11}{3} \) with a slope of \(-\frac{2}{3}\) also passing through \((-1, -3)\).

Conclusion: Tangent and Normal Line Equations

The tangent line equation at \((-1, -3)\) is \( y = \frac{3}{2}x - \frac{3}{2} \) and the normal line equation is \( y = -\frac{2}{3}x - \frac{11}{3} \).

Key concepts

Parabola

A parabola is a U-shaped curve that can open either up, down, left, or right, depending on how its equation is structured. In our case, the equation is given by \( y^2 = -9x \). Since the \( y \) term is squared and the coefficient of \( x \) is negative, this parabola opens to the left.

This equation doesn't represent the typical "up" or "down" facing parabolas (like \( y = ax^2 \)). Instead, it features a focus and directrix, which are used to define the parabola in terms of a set of points. The vertex of our parabola is at the origin \((0,0)\).
We can visualize this particular parabola as encircling the y-axis and opening towards the negative x-axis.

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are mixed together. In the given parabola equation \( y^2 = -9x \), both \( y \) and \( x \) appear together, not separated by a function \( y = f(x) \).

To find the tangent's slope, we implicitly differentiate both sides with respect to \( x \). By applying the chain rule to the left side, \( 2y \frac{dy}{dx} \), we recognize the need to multiply by \( \frac{dy}{dx} \) since \( y \) itself depends on \( x \).
As a result, the equation becomes \( 2y \frac{dy}{dx} = -9 \), allowing us to solve for \( \frac{dy}{dx} \) and find the slope of the tangent at a specific point.

Slope of a Line

The slope of a line is a measure of its steepness or inclination. It's typically expressed as a fraction \( \frac{rise}{run} \) or as the change in \( y \) over the change in \( x \) between two points on the line.

In our example, the slope of the tangent line of the parabola at the point \((-1, -3)\) is calculated to be \( \frac{3}{2} \). This means that for every 2 units the line moves horizontally, it moves 3 units vertically. A positive slope usually indicates an upward trend from left to right.
Meanwhile, the slope of the normal line, which is perpendicular to the tangent, is found by taking the negative reciprocal of the tangent's slope, resulting in \(-\frac{2}{3}\). This perpendicular slope confirms that the normal line generally moves in an opposite direction compared to the tangent.

Point-Slope Form

The point-slope form is a linear equation format that uses a point on the line and the slope to define the line. It's particularly useful when you need to construct a line quickly given a point and the line's slope.

For a line with slope \( m \) passing through point \((x_1, y_1)\), the point-slope form is given by:
\[ y - y_1 = m(x - x_1) \]
This form is perfect for writing equations of tangent and normal lines.
In the solution, for the tangent line at \((-1, -3)\) with a slope of \( \frac{3}{2} \), we use the point-slope form to write:
\[ y + 3 = \frac{3}{2} (x + 1) \]
Similarly, the normal line, utilizing \( -\frac{2}{3} \) as its slope, is expressed as:
\[ y + 3 = -\frac{2}{3} (x + 1) \]