Question

Find the equation of the line tangent to the graph of \(x^{2}+\) \(y^{2}+x y=7\) at the point (1,2).

Short answer

The tangent line equation is \(y = -\frac{4}{5}x + \frac{9}{5}.\)

Step-by-step solution

Understand the Problem

We need to find the equation of the tangent line to the curve given by the equation \(x^2 + y^2 + xy = 7\) at the specific point \((1,2)\). To find the tangent line, we'll calculate the slope of the tangent line at this point and use the point-slope form of a line equation.

Implicit Differentiation

Since the equation of the graph is given implicitly, we'll use implicit differentiation to find \(\frac{dy}{dx}\). Differentiate both sides of the equation \(x^2 + y^2 + xy = 7\) with respect to \(x\): \[ 2x + 2y\frac{dy}{dx} + y + x\frac{dy}{dx} = 0. \] Reorganize to solve for \(\frac{dy}{dx}\).

Solve for the Derivative

Continuing from the differentiation: \[ 2y\frac{dy}{dx} + x\frac{dy}{dx} = -2x - y \] Factor out \(\frac{dy}{dx}\): \[ \frac{dy}{dx}(2y + x) = -2x - y. \] Thus, \[ \frac{dy}{dx} = \frac{-2x - y}{2y + x}. \]

Evaluate the Derivative at Point (1,2)

Plug \(x = 1\) and \(y = 2\) into the derivative: \[ \frac{dy}{dx} \bigg|_{(1,2)} = \frac{-2(1) - 2}{2(2) + 1} = \frac{-4}{5}. \] The slope of the tangent line at the point \((1,2)\) is \(-\frac{4}{5}\).

Write the Equation of the Tangent Line

Use the point-slope form of a line with point \((1,2)\) and slope \(-\frac{4}{5}\): \[ y - 2 = -\frac{4}{5}(x - 1). \] Simplify to get the equation of the tangent line: \[ y = -\frac{4}{5}x + \frac{9}{5}. \]

Key concepts

Implicit Differentiation

When we have an equation involving both the variables \(x\) and \(y\), and we can't easily solve for \(y\) in terms of \(x\), we use implicit differentiation. This method lets us find the derivative \(\frac{dy}{dx}\) even if \(y\) isn't isolated. Here’s how it works:
- Differentiate each term with respect to \(x\).- Apply the chain rule to terms involving \(y\). Treat \(y\) as a function of \(x\), so its derivative becomes \(\frac{dy}{dx}\).- Collect all terms involving \(\frac{dy}{dx}\) on one side of the equation.- Solve for \(\frac{dy}{dx}\).For example, with the equation \(x^2 + y^2 + xy = 7\), differentiate to get:1. The derivative of \(x^2\) is \(2x\).2. The derivative of \(y^2\) is \(2y \frac{dy}{dx}\).3. The derivative of \(xy\) is \(y + x \frac{dy}{dx}\) using the product rule.This gives us an equation to solve for \(\frac{dy}{dx}\), which is essential for finding the slope of the tangent.

Derivative

The derivative \(\frac{dy}{dx}\) represents the rate at which \(y\) changes with respect to \(x\). It's crucial for understanding how functions behave over time or space, especially when finding tangent lines.

Here's why it's important when solving our problem:- We found the derivative \(\frac{dy}{dx} = \frac{-2x - y}{2y + x}\). This formula tells us the slope of the tangent line at any point on the curve.- By substituting \(x = 1\) and \(y = 2\) into the derivative formula, we computed the slope = \(-\frac{4}{5}\).
Like calculating speed from distance and time, derivatives help us find the slope, representing the tangent line's tilt at a specific point. The steeper the slope, the more tilted the line is.

Point-Slope Form

Once we know the slope of a tangent line, the point-slope form is a powerful tool to write the line's equation.The point-slope form of a line is given by: \[y - y_1 = m(x - x_1)\]where \((x_1, y_1)\) is our point, and \(m\) is the slope.
In our case:- The point is \((1, 2)\).- The slope \(m\) is \(-\frac{4}{5}\) from our derivative calculation.Plug these values into the formula to get:\[y - 2 = -\frac{4}{5}(x - 1)\]This equation tells us how \(y\) changes for every change in \(x\) based on the slope. It's a straightforward way to get from knowing the slope and a point to the line equation.
Finally, simplifying this equation, we convert it into its slope-intercept form \(y = mx + c\) for convenience, getting \(y = -\frac{4}{5}x + \frac{9}{5}\). This brings out the equation's information ready for practical use or graphing.