Question

(a) Find an equation of the normal line to the ellipse \(\frac{x^{2}}{32}+\frac{y^{2}}{8}=1\) at the point \((4,2) .(\text { b) Use a graphing }\) utility to graph the ellipse and the normal line. (c) At what other point does the normal line intersect the ellipse?

Short answer

The equation of the normal line to the ellipse at the point (4,2) is \(y = 8x - 30\). Also, the other point of intersection between the normal line and the ellipse can be found by solving the equation \(\frac{x^{2}}{32} + \frac{(8x-30)^2}{8} = 1 \), to get the x and y coordinates of that point.

Step-by-step solution

Find the derivative of the given equation

The derivative of the equation \(\frac{x^{2}}{32}+\frac{y^{2}}{8}=1\) with respect to \(x\) can be found using the power rule: \(y'(x)=\frac{-x}{16y}\). This gives the slope of the tangent line at any point on the ellipse.

Find the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Hence, if the slope of the tangent line is \(y'(4,2)=\frac{-4}{16*2}=\frac{-1}{8}\), the slope of the normal line is, \(m=-\frac{1}{y'(4,2)}=8\).

Find the equation of the normal line

Using the point-slope form of the line equation, \(y - y1 = m (x - x1)\), where (x1, y1) = (4,2), and m = the slope of the normal line (8), the equation becomes: \(y - 2 = 8 (x - 4)\), which simplifies to \(y = 8x - 30\). This is the equation of the normal line at point (4,2).

Find the other point of intersection between the normal line and the ellipse

Substitute the equation of line, \(y = 8x - 30\), into the equation of the ellipse. The resulting equation is: \(\frac{x^{2}}{32} + \frac{(8x-30)^2}{8} = 1 \). Solve this quadratic equation to find the x-values at the point of intersection. The points of intersection will have \(y\) values that correspond to the \(x\) values found, using the equation of the line.

Key concepts

Tangent and Normal Lines

Understanding the relationship between tangent and normal lines to a curve is a fundamental concept in calculus and geometry. A tangent line touches a curve at a single point without intersecting it, and its slope at that point reflects the curve's slope. On the other hand, a normal line is perpendicular to the tangent line at the point of tangency. This perpendicularity means that the slope of the normal line is the negative reciprocal of the slope of the tangent line.

For our ellipse \(\frac{x^{2}}{32}+\frac{y^{2}}{8}=1\), finding a normal line at a specified point involves calculating the slope of the tangent line at that point first. If the slope of the tangent line is \(m\), then the slope of the normal line will be \(\frac{-1}{m}\). This geometric relationship is crucial when analyzing curves and their properties, as we often want to understand how curves behave locally in the vicinity of a given point.

Calculating slopes of tangent and normal lines also has practical applications, such as in physics for understanding motion, or in engineering for stress analysis.

Point-Slope Form

The point-slope form of a line is an essential tool in coordinate geometry and is expressed as \(y - y1 = m(x - x1)\), where \(m\) is the slope of the line, and \(x1, y1)\) is a known point on the line. This form is particularly useful for quickly writing down the equation of a line when you know a point it passes through and its slope.

In our exercise, we apply the point-slope form to find the equation of the normal line. With the point (4,2) on the ellipse and knowing the slope of the normal line is 8, we use point-slope form directly to produce \(y - 2 = 8(x - 4)\). After rearranging, we get \(y = 8x - 30\), representing the normal line's equation. The simplicity of point-slope form can be especially appreciated when dealing with problems that require the equations of tangent or normal lines to curves.

Derivative of Implicit Functions

The derivative of implicit functions is a slightly trickier concept in calculus, mainly because the function is not explicitly solved for one variable in terms of another. In our given ellipse's equation, \(\frac{x^{2}}{32}+\frac{y^{2}}{8}=1\), neither \(x\) nor \(y\) is isolated. Here, we find the derivative implicitly, where we differentiate both sides of the equation with respect to \(x\) while treating \(y\) as a function of \(x\).

This process involves applying the chain rule for differentiating compositions of functions, where we end up with the derivative of \(y\) with respect to \(x\), denoted as \(y'\). In our case, we find \(y'\) to be \(\frac{-x}{16y}\), which lets us determine the slope of the tangent line at any point on the ellipse. Differentiation of implicit functions is a valuable technique when the relationship between variables cannot be explicitly separated and has wide-ranging applications across multiple fields of study.

Quadratic Equations

Quadratic equations, which are polynomials of degree two, typically take the form \(ax^2 + bx + c = 0\). They can represent various phenomena such as the trajectories of projectiles, the optimization of areas, and the behavior of certain financial models. In every quadratic equation, the highest exponent is two, which means the graph of a quadratic function is a parabola.

During our exercise, we encounter a quadratic equation when we substitute the equation of the normal line into the ellipse's equation. This procedure transforms the relationship into a quadratic in terms of \(x\), which we can then solve using conventional methods like factoring, completing the square, or applying the quadratic formula. By solving the quadratic equation, we find the other point(s) where the normal line intersects with the ellipse — an invaluable outcome when assessing the geometry of curves and their interactions with lines.

Understanding how to solve quadratic equations is essential as it is a frequent requisite in not just geometry, but in algebra and calculus problems as well.