Problem 79 - Calculus for Biology and Medicine

Question

Find all tangent lines to the curve $$ y=x^{2} $$ that pass through the point \(\left(0,-a^{2}\right)\), where \(a\) is a positive number.

Step-by-Step Solution

Answer

The tangent lines are \(y = 2ax - a^2\) and \(y = -2ax - a^2\).

Step 1: Understand the problem

We need to find tangent lines to the curve \(y = x^2\) that pass through the point \((0, -a^2)\). A tangent to \(y = x^2\) at a point \((c, c^2)\) has the equation \(y = 2cx - c^2\).

Step 2: Set up the tangent line equation

The tangent line will be of the form \(y = 2cx - c^2\), where \(c\) is the x-coordinate of the point of tangency on the curve. We want this line to pass through \((0, -a^2)\).

Step 3: Substitute point into tangent line equation

Substitute \((0, -a^2)\) into the equation \(y = 2cx - c^2\) to get \(-a^2 = 2c \cdot 0 - c^2\), which simplifies to \(-a^2 = -c^2\) or \(c^2 = a^2\).

Step 4: Solve for the point of tangency

From \(c^2 = a^2\), we find \(c = a\) or \(c = -a\). Thus, the points of tangency on the curve are \((a, a^2)\) and \((-a, a^2)\).

Step 5: Write equations of the tangent lines

Using the point \((a, a^2)\), the tangent line is \(y = 2ax - a^2\). For the point \((-a, a^2)\), the tangent line is \(y = -2ax - a^2\).

Key Concepts

Derivative

The concept of a derivative is fundamental in calculus. It provides us with a way to understand how a function changes at any given point. In simple terms, the derivative at a point on a curve gives the slope of the tangent line to the curve at that point. For a function like \( y = x^2 \), the derivative with respect to \( x \) is computed using the power rule. To find the derivative of \( y = x^2 \), we use the formula for derivatives of powers of \( x \), which states that the derivative of \( x^n \) is \( nx^{n-1} \). Applying this rule:

• The derivative of \( x^2 \) is \( 2x \).

This derivative, \( 2x \), represents the slope of the tangent line to the curve \( y = x^2 \) at any point \( x \). Knowing how to compute and interpret derivatives is crucial when determining equations of tangent lines.

Curve Equation

The curve equation represents the relationship between dependent and independent variables on a graph. For this problem, we are looking at the curve given by \( y = x^2 \).This type of equation is called a quadratic equation, and it forms a parabola when graphed. Understanding the shape and behavior of the curve is essential, as:

• A parabola opens upwards if the coefficient of \( x^2 \) is positive, as it is in this case.
• The lowest point of the parabola is called the vertex, located at the origin \((0, 0)\) for \( y = x^2 \).

Knowing these characteristics of the curve helps us visualize where the tangent lines will touch, or "be tangent to," the curve. It's at these points of tangency where the tangent line will have the same slope as the curve gradient, dictated by its derivative.

Point of Tangency

The point of tangency is where the tangent line meets the curve. This is a vital concept because it solidifies the relationship between a curve and its specific tangent line.For the exercise involving \( y = x^2 \):

• The tangent line equation at a point \( (c, c^2) \) is given by \( y = 2cx - c^2 \).
• We find the values of \( c \) by setting up the line to also pass through the specified external point \((0, -a^2)\).
• By solving \( c^2 = a^2 \), the x-coordinates of the points of tangency are identified as \( c = a \) and \( c = -a \).

These coordinates \((a, a^2)\) and \((-a, a^2)\) on the curve are crucial as they denote exactly where the tangent lines meet the curve without cutting through it.

Substitution Method

The substitution method is a technique used to solve systems of equations or to simplify complex expressions in mathematical problems. In our exercise, it plays a significant role in determining the points of tangency.Examining the given problem:

• The goal is to find where the tangent line to the curve passes through a specific point, \((0, -a^2)\).
• Substitute this point into the tangent line equation \( y = 2cx - c^2 \) to get \(-a^2 = -c^2 \).
• Solving \(-a^2 = -c^2\) gives \( c^2 = a^2 \), leading to two possible solutions for \( c \), which are \( c = a \) and \( c = -a \).

Through substitution, these solutions are found efficiently, providing us with the specific coordinates for the points of tangency. The substitution method simplifies the problem, reducing it to easily manageable components and leading to solutions that define the tangent lines.