Chapter 4: Problem 79
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.
Short Answer
Expert verified
The tangent lines are \(y = 2ax - a^2\) and \(y = -2ax - a^2\).
Step by step solution
01
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 \(y = 2cx - c^2\).
02
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 \((0, -a^2)\).
03
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 \(c^2 = a^2\).
04
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 \((-a, a^2)\).
05
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 \(y = -2ax - a^2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( 2x \).
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 \( y = x^2 \).
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 \( c = -a \).
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 \( c = -a \).