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. $$ y^{2}=16 x,(1,-4) $$

Short answer

Tangent line: \( y = -2x + 2 \), Normal line: \( y = \frac{1}{2}x - \frac{9}{2} \).

Step-by-step solution

Understand the parabola equation

The equation of the parabola is given as \( y^2 = 16x \). This is a sideways parabola that opens to the right.

Find the derivative to get the slope of the tangent

To find the slope of the tangent line, differentiate \( y^2 = 16x \) implicitly. Start by differentiating both sides with respect to \( x \):\[ 2y \frac{dy}{dx} = 16 \]From this, solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{16}{2y} = \frac{8}{y} \]

Compute the slope of the tangent

Substitute the given point \( (1, -4) \) into the derivative to find the slope of the tangent line:\[ \frac{dy}{dx} \bigg|_{(1, -4)} = \frac{8}{-4} = -2 \]

Write the equation of the tangent line

Use the point-slope form of the line equation to write the tangent line:\[ y - y_1 = m (x - x_1) \] \Where \( m \) is the slope and \( (x_1, y_1) \) is the point. Plugging in \( m = -2 \), \( x_1 = 1 \), and \( y_1 = -4 \):\[ y + 4 = -2 (x - 1) \]Simplifying, the equation becomes:\[ y = -2x + 2 \]

Find the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Given that the tangent slope is \(-2\), the normal line slope is:\[ m_{normal} = \frac{1}{2} \]

Write the equation of the normal line

Use the point-slope form again for the normal line:\[ y - y_1 = m_{normal} (x - x_1) \]Substitute \( m_{normal} = \frac{1}{2} \), \( x_1 = 1 \), \( y_1 = -4 \):\[ y + 4 = \frac{1}{2} (x - 1) \]This simplifies to:\[ y = \frac{1}{2}x - \frac{9}{2} \]

Sketch the parabola, tangent, and normal lines

Draw the sideways parabola \( y^2 = 16x \). Mark the point \( (1, -4) \). Sketch the tangent line \( y = -2x + 2 \) passing through \( (1, -4) \), and draw the normal line \( y = \frac{1}{2}x - \frac{9}{2} \) that is perpendicular to the tangent at that same point.

Key concepts

Parabola

A parabola is a well-known curve that can be defined by a simple equation. In this exercise, we deal with the equation \( y^2 = 16x \), which represents a specific type of parabola. Unlike the more common upward or downward opening parabolas, this one opens sideways.
This parabola opens to the right because the \( y^2 \) term is isolated, and the \( x \) term is positive. The standard form for a sideways parabola is \((y-k)^2 = 4p(x-h)\), where \((h, k)\) represents the vertex, and \(p\) is the distance from the vertex to the focus. Here, the equation can be rewritten as \((y-0)^2 = 4(4)(x-0)\), so the vertex is at \((0, 0)\) and the focus is at \((4, 0)\).

• **Vertex**: The point where the parabola changes direction. Here, it's \((0, 0)\).
• **Focus**: A point inside the parabola that helps define its exact shape. Here, it's located at \((4, 0)\).

Understanding the structure of the parabola helps in sketching it and analyzing its geometry for problems involving tangent and normal lines.

Implicit Differentiation

Implicit differentiation is a powerful technique used to differentiate equations not easily solved for one variable. It comes in handy when dealing with equations involving two or more variables like the one provided here.
For the equation \( y^2 = 16x \), solving explicitly for \(y\) can be cumbersome, so we use implicit differentiation instead. Applying it, we differentiate both sides with respect to \(x\). On the left side, for \(y^2\), we use the chain rule, considering \(y\) as a function of \(x\):
\[ 2y \frac{dy}{dx} = 16 \]
This partial differentiation allows us to express the slope \(\frac{dy}{dx}\), representing the rate at which \(y\) changes with respect to \(x\).
By rearranging the equation, we isolate \(\frac{dy}{dx}\) to find the slope:
\[ \frac{dy}{dx} = \frac{16}{2y} = \frac{8}{y} \]
This derivative gives us the slope of the tangent line at any point on the parabola. A crucial step in finding the tangent and normal lines to a particular point.

Slope Calculation

Finding the slope of a line is crucial in determining both tangent and normal lines. For this exercise, calculating the tangent line's slope involves substituting the given point into the derivative.
Given the derivative \( \frac{dy}{dx} = \frac{8}{y} \), we substitute the point \((1, -4)\):
\[ \frac{dy}{dx} \bigg|_{(1, -4)} = \frac{8}{-4} = -2 \]
This value gives us the slope of the tangent line at that specific point.
For the normal line, we need the negative reciprocal of the tangent slope because the normal is perpendicular to the tangent. If a line has slope \(m\), a line perpendicular to it will have a slope \(-\frac{1}{m}\).
Using the tangent slope of \(-2\), the normal line's slope becomes:
\[ m_{normal} = \frac{1}{2} \] Understanding how to find these slopes ensures you can write the correct equations for both lines.

Line Equations

Line equations provide a way to express straight lines mathematically, and they are determined using point-slope form when the slope and one point on the line are known.
The point-slope formula is given by:

• \( y - y_1 = m(x - x_1) \)

Here, \(m\) is the slope, and \((x_1, y_1)\) is a specific point on the line.
Using our slope for the tangent line \(-2\) and point \((1, -4)\), the equation becomes:
\[ y + 4 = -2(x - 1) \]
Simplifying gives us:
\[ y = -2x + 2 \]
For the normal line, with slope \(\frac{1}{2}\) and the same point \((1, -4)\), we have:
\[ y + 4 = \frac{1}{2}(x - 1) \]
This simplifies to:
\[ y = \frac{1}{2}x - \frac{9}{2} \]
Drawing these lines can visually show their relationship: the tangent line just touches the parabola at \((1, -4)\), while the normal line cuts across and is perpendicular to the tangent.