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. $$ x^{2}=2 y,(4,8) $$

Short answer

Tangent line: \(y = 4x - 8\); Normal line: \(x + 4y = 36\).

Step-by-step solution

Differentiate the Given Parabola

The parabola is given by the equation \(x^2 = 2y\). First, we need to differentiate this equation with respect to \(x\) to find the slope of the tangent line. Differentiating both sides with respect to \(x\), we have: \[\frac{d}{dx}(x^2) = \frac{d}{dx}(2y)\] This simplifies to: \[2x = 2\frac{dy}{dx}\] Therefore, \(\frac{dy}{dx} = x\).

Find the Slope at the Given Point

We need to find the slope of the tangent line at the point (4,8). Use the derivative calculated in Step 1:\[\frac{dy}{dx} = x\] Substituting \(x = 4\),\[\text{slope of tangent line} = \frac{dy}{dx} = 4\].

Write the Equation of the Tangent Line

With the slope of the tangent line known (4) and the point (4,8), we use the point-slope form of a line: \[y - y_1 = m(x - x_1)\] Substituting \(m = 4\), \(x_1 = 4\), and \(y_1 = 8\), we have: \[y - 8 = 4(x - 4)\] Which simplifies to: \[y = 4x - 8\].

Find the Slope of the Normal Line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Thus, if the slope of the tangent is 4, the slope of the normal line is \(-\frac{1}{4}\).

Write the Equation of the Normal Line

Using the slope of the normal line \(-\frac{1}{4}\) and the point (4,8), we apply the point-slope form again: \[y - 8 = -\frac{1}{4}(x - 4)\] Simplifying:\[4(y - 8) = -(x - 4)\] \[4y - 32 = -x + 4\] \[x + 4y = 36\].

Sketch the Graph

Plot the parabola \(x^2 = 2y\), which is a sideways parabola opening upward. The vertex is at the origin \((0,0)\). Then, plot the tangent line \(y = 4x - 8\) and the normal line \(x + 4y = 36\) at the point (4,8), ensuring the tangent line has a slope of 4 and the normal line has a slope of \(-\frac{1}{4}\) at the given point.

Key concepts

Calculus

Calculus is a branch of mathematics that focuses on understanding change and motion. It primarily involves two core concepts: differentiation and integration. Differentiation concerns finding the rate at which things change, while integration deals with finding the total or accumulative effect. In this particular problem, we use differentiation to find the slope of tangent and normal lines to a parabola. The process of setting a small change in the equation helps to calculate derivatives, which lays the groundwork for finding these slopes. Calculus is essential in many fields, from physics to economics, thanks to its ability to model and analyze continuous change.

Derivatives

Derivatives provide a way to calculate the slope of a tangent to a curve at any given point. They represent the rate of change of a function. In simple terms, if you have a function that describes a curve, the derivative tells you how steep that curve is at any point. For our exercise, the parabola equation is given by \(x^2 = 2y\). By differentiating both sides with respect to \(x\), we find the derivative \(\frac{dy}{dx} = x\). This derivative gives us the slope of the tangent line, which is essential to find tangents and normals on the curve.

Parabolas

Parabolas are unique U-shaped curves that are symmetric around an axis and open either upward or downward, or sideways based on their equation structure. The equation \(x^2 = 2y\) represents a sideways parabola that opens upward. Parabolas are two-dimensional shapes that play an important role in various real-world applications, such as satellite dish designs and headlight reflectors, due to their reflective properties. Their structure allows them to focus points, which geometrically explains phenomena like the path of projectiles under uniform acceleration.

Slope

Slope is a measure of steepness or the angle of a line with respect to the horizontal axis. In the context of curves and calculus, the slope of a tangent line to a curve at a point provides an instantaneous rate of change of the curve. For linear equations, determining the slope involves finding the ratio of the vertical change to the horizontal change between two points, often described as 'rise over run'. In our exercise, we've computed the slope of the tangent line as \(4\) and the normal line as \(-\frac{1}{4}\). The negative reciprocal indicates the perpendicular nature of the normal to the tangent.

Analytical Geometry

Analytical geometry, also known as coordinate geometry, is the study of geometry using a coordinate system. This approach allows the solving of geometric problems algebraically. It combines algebra and geometry to find properties of geometric shapes. In this exercise, we used point-slope form to derive the equations of the tangent and normal lines. Analytical geometry enables us to deliver precise calculations for curves and lines by leveraging equations and coordinates, facilitating an analytical approach to geometric problems.