Question

Find the equation of the tangent line to the given curve at the given point. $$ \frac{x^{2}}{27}+\frac{y^{2}}{9}=1 \text { at }(3,-\sqrt{6}) $$

Short answer

The equation of the tangent line is \( x - \sqrt{6}y = 9.\)

Step-by-step solution

Identify the Curve

The equation \( \frac{x^2}{27} + \frac{y^2}{9} = 1 \) represents an ellipse centered at the origin. The x-term is divided by 27, and the y-term is divided by 9.

Find the Derivative

To find the slope of the tangent line, we need to differentiate the equation of the curve. Start by implicitly differentiating the equation: \[ \frac{d}{dx}\left( \frac{x^2}{27} + \frac{y^2}{9} \right) = \frac{d}{dx}(1). \] This gives us: \[ \frac{2x}{27} + \frac{2y}{9} \cdot \frac{dy}{dx} = 0. \] Solving for \( \frac{dy}{dx} \), we get \[ \frac{dy}{dx} = -\frac{9x}{27y} = -\frac{x}{3y}. \]

Substitute the Point into the Derivative

Substitute the point \((3, -\sqrt{6})\) into the derivative to find the slope of the tangent line: \[ \frac{dy}{dx} \bigg|_{(3, -\sqrt{6})} = -\frac{3}{3(-\sqrt{6})} = \frac{1}{\sqrt{6}}. \] Therefore, the slope of the tangent line at the point is \( \frac{1}{\sqrt{6}}. \)

Find the Equation of the Tangent Line

Using the point-slope form of the equation \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point, we substitute \( m = \frac{1}{\sqrt{6}} \) and \( (x_1, y_1) = (3, -\sqrt{6}) \): \[ y + \sqrt{6} = \frac{1}{\sqrt{6}} (x - 3). \] Simplify to find: \[ y = \frac{1}{\sqrt{6}} x - \frac{3}{\sqrt{6}} - \sqrt{6}. \]

Simplify the Equation of the Tangent Line

Multiply every term by \( \sqrt{6} \) to eliminate the radical: \[ \sqrt{6}y = x - 3 - 6, \] or simplified, \[ \sqrt{6}y = x - 9. \] Thus, the equation of the tangent line is \[ x - \sqrt{6}y = 9. \]

Key concepts

Ellipse

An ellipse is a geometric shape that looks like a squashed or elongated circle. In mathematics, an ellipse is defined by an equation of the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, the constants \( a \) and \( b \) represent the semi-major and semi-minor axes, respectively. These axes determine the shape and size of the ellipse.
Ellipses can be found in many real-world applications, such as planetary orbits and optical systems.

• When \( a > b \), the ellipse is longer horizontally.
• When \( b > a \), it is longer vertically.

The given problem involves an ellipse centered at the origin, with semi-major axis \( a = \sqrt{27} \) and semi-minor axis \( b = 3 \). This orientation naturally influences how we calculate slopes and draw tangents to the curve.

Implicit Differentiation

Implicit differentiation is a mathematical technique used to find the derivative of functions that are not explicitly solved for one variable in terms of the other. Often used with equations that define shapes like ellipses, this method saves us the trouble of solving for \( y \) first.
For the ellipse given by \( \frac{x^2}{27} + \frac{y^2}{9} = 1 \), we cannot express \( y \) directly as a function of \( x \) without complicating matters. By differentiating both sides with respect to \( x \), including terms that involve \( y \), we can find \( \frac{dy}{dx} \) just as easily.
This involves:

• Using the chain rule, as \( y \) is considered an implicit function of \( x \).
• Treating \( \frac{dy}{dx} \) as a derivative term whenever \( y \) is differentiated.

The solution to our original exercise demonstrates this method to find the slope, \( \frac{dy}{dx} = -\frac{x}{3y} \), which we then evaluate at a specific point.

Point-Slope Form

The point-slope form of a linear equation is useful for finding the equation of a line given a point and the slope. This form is written as \( y - y_1 = m(x - x_1) \), where:

• \( m \) represents the slope of the line.
• \( (x_1, y_1) \) is a specific point on the line.

In our problem, after determining the slope of the tangent line \( m = \frac{1}{\sqrt{6}} \) at the point \( (3, -\sqrt{6}) \), we apply this formula.
By substituting the point and slope into the equation, we initially obtain \( y + \sqrt{6} = \frac{1}{\sqrt{6}} (x - 3) \).
This allows us to express the tangent line in a clear and straightforward way, making it easier to visualize and understand the line's behavior on the graph relative to the ellipse.

Slope of the Tangent Line

The slope of the tangent line to a curve at a specific point is a crucial concept in calculus. It represents the rate at which the curve is changing at that exact location.
For the given ellipse, we determined the derivative \( \frac{dy}{dx} = -\frac{x}{3y} \), which helps us calculate this slope. By evaluating this expression at the point \( (3, -\sqrt{6}) \), we find the slope to be \( \frac{1}{\sqrt{6}} \).
To understand why this is useful:

• The slope tells us whether the tangent is rising, falling, or flat at the point of tangency.
• It helps in constructing the equation of the tangent line.

By knowing the slope and the point, using the point-slope form, we can construct the equation of the tangent that is needed for further analysis or graphical representation.