Question

Let \(L\) be a normal to the parabola \(y^{2}=4 x\). If \(L\) passes through the point \((9,6)\), then \(L\) is given by (A) \(y-x+3=0\) (B) \(y+3 x-33=0\) (C) \(y+x-15=0\) (D) \(y-2 x+12=0\)

Short answer

The normal line to the parabola \(y^{2}=4 x\) that passes through the point \((9,6)\) is given by the equation \(y + 3x - 33 = 0\), which is option (B).

Step-by-step solution

Determine the Slope of the Tangent Line

The slope of the tangent line to the parabola where the normal line intersects can be found using the derivative of the parabola's equation. Given the parabola equation, we can express it as a function: \(y^2 = 4x \implies y^2 = 4f(x)\), where \(f(x) = x\). Now we differentiate \(y^2\) with respect to \(x\) to find the slope of the tangent line: \(2y\frac{dy}{dx} = 4\). Solving for \(\frac{dy}{dx}\), we find that the slope of the tangent line is \(\frac{2}{y}\).

Find the Slope of the Normal Line

The slope of the normal line to the curve is the negative reciprocal of the slope of the tangent line. Knowing the slope of the tangent is \(\frac{2}{y}\), we find the slope of the normal line, which we'll call \(m_n\), by taking \(m_n = -\frac{y}{2}\).

Identify the Point of Intersection

We know the normal line passes through the point \((9,6)\). We can substitute \(y = 6\) in the equation \(y^2 = 4x\) to find the associated x-coordinate on the parabola, which is the x-coordinate of the point of intersection: \(6^2 = 4x\) leads to \(x = 9\). Hence, our point of intersection on the parabola is also \((9,6)\), and the x and y coordinates are consistent.

Calculate the Slope of the Normal Line at the Intersection

Use the y-coordinate of the point at which the normal line intersects the parabola (which is 6) to find the slope of the normal line using the formula from Step 2: \(m_n = -\frac{y}{2} = -\frac{6}{2} = -3\).

Write the Equation of the Normal Line Using Point-Slope Form

With the slope of the normal line, \(m_n = -3\), and the point of intersection, \((9, 6)\), use the point-slope form of the line equation to write the equation of the normal line: \(y - y_1 = m_n(x - x_1)\), where \((x_1, y_1)\) is the point \((9, 6)\). This yields \(y - 6 = -3(x - 9)\).

Simplify the Equation of the Normal Line

Expand and rearrange the equation from Step 5 to get it into the general form \(y = mx + b\): \(y - 6 = -3x + 27\). Adding 6 to both sides, we obtain the standard form linear equation: \(y = -3x + 33\).

Confirm the Correct Answer Choice

Compare the standard form equation we derived with the multiple-choice options to find a match. The linear equation that represents the normal line is: \(y + 3x - 33 = 0\), which corresponds to option (B).

Key concepts

Differential Calculus

Differential calculus is a subfield of calculus concerned with the study of how things change. It allows us to determine the rate at which quantities change. One fundamental tool in differential calculus is the derivative. The derivative of a function at a point is a measure of the rate at which the function's value is changing at that particular point. It is often represented as \( \frac{dy}{dx} \) or \( f'(x) \).

In the context of the parabola \( y^2 = 4x \), the process of finding the slope of the tangent line involves taking the derivative of the parabola's equation with respect to \( x \). This is calculated by differentiating both sides of the equation, resulting in \( 2y\frac{dy}{dx} = 4 \) and solving for \( \frac{dy}{dx} \), which represents the slope of the tangent line at any point on the parabola. Understanding how to take derivatives is essential for solving many problems in calculus, physics, and other sciences.

Slope of Tangent and Normal

In calculus, the concept of slope is pivotal when discussing lines that are tangent or normal to a curve. The slope of a tangent line at a given point on a curve is the rate at which the curve is changing at that point. If the curve is represented by a function \( y = f(x) \), the slope at point \( P \) is the value of the derivative \( \frac{dy}{dx} \) at \( P \).

The normal line is perpendicular to the tangent line at the point of tangency. To find the slope of the normal line, \( m_n \)—also called the normal slope—we take the negative reciprocal of the slope of the tangent line. If the slope of the tangent is \( m_t \), then the slope of the normal line is \( m_n = -\frac{1}{m_t} \). Knowing how to find the slopes of tangent and normal lines allows us to accurately describe the geometry of curves and is essential for solving problems related to rates of change and optimization.

Point-Slope Form Equation

The point-slope form equation is a straightforward method for writing the equation of a line when you know the slope and a point through which the line passes. The general form of this equation is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope of the line and \( (x_1, y_1) \) are the coordinates of the given point.

When writing the equation of a normal line to a parabola, as in our example, we first determine the slope of the normal line, \( m_n \) and then use the point-slope form with the known point of intersection on the parabola, effectively plugging these values into the equation to find the specific linear equation that represents the normal. This form is particularly useful for writing equations quickly and is often used as a step towards deriving other forms of the line equation, such as the slope-intercept form or standard form.