91Ó°ÊÓ

Implicit differentiation is a technique used when you have an equation where both variables, like \(x\) and \(y\), are tangled together, as seen in our ellipse equation \(x^2 + 3y^2 = 76\). In such cases, solving for one variable in terms of another isn't straightforward. Instead, we differentiate both sides with respect to one variable—usually \(x\).
Here's how it works: When you come across a term like \(y^2\), you treat \(y\) as a function of \(x\) and apply the chain rule. This is why the differentiation of \(y^2\) becomes \(2y \frac{dy}{dx}\). By applying implicit differentiation to our ellipse equation, we differentiate each term and acquire:

• The derivative of \(x^2\) is \(2x\)
• The derivative of \(3y^2\) is \(6y \frac{dy}{dx}\)

Combining these, we get \(2x + 6y \frac{dy}{dx} = 0\). Solving for \( \frac{dy}{dx} \) gives us the slope of the tangent line.

The slope of a tangent line is a crucial part of finding the equation of a tangent. It tells us the direction and steepness of the line at the point it touches the ellipse. Once we have our derivative from implicit differentiation, \( \frac{dy}{dx} = -\frac{x}{3y} \), we can substitute the specific \(x\) and \(y\) values of the point on the ellipse to find the slope \(m\).
For each of the points:\

• At \((8, 2)\), substituting gives \(m = -\frac{8}{3 \times 2} = -\frac{4}{3}\)
• At \((-7, 3)\), substituting gives \(m = -\frac{-7}{3 \times 3} = \frac{7}{9}\)
• At \((1, -5)\), substituting gives \(m = -\frac{1}{3 \times (-5)} = \frac{1}{15}\)

Each slope is specific to the point on the ellipse where the tangent touches. This makes the slope essential for the next step of writing the tangent line equation.

Once we know the slope \(m\) of the tangent line and the point \((x_1, y_1)\) where it touches the ellipse, we can write the equation using the point-slope form \(y - y_1 = m(x - x_1)\). This forms the core of writing the tangent equation.
For example:

• Using \((8, 2)\) and \(m = -\frac{4}{3}\), we get:\
  \[y - 2 = -\frac{4}{3}(x - 8)\] which simplifies to \[y = -\frac{4}{3}x + \frac{38}{3}\].
• Using \((-7, 3)\) and \(m = \frac{7}{9}\), we have: \
  \[y - 3 = \frac{7}{9}(x + 7)\] simplifying to \[y = \frac{7}{9}x + \frac{76}{9}\].
• Using \((1, -5)\) and \(m = \frac{1}{15}\), we get:
  \[y + 5 = \frac{1}{15}(x - 1)\] simplifying to \[y = \frac{1}{15}x - \frac{76}{15}\].

Writing each tangent's equation in \(y = mx + b\) form allows us to easily identify the slope and the y-intercept \(b\) of the line.