Standard Form of the Ellipse

Oct 13, 2025 by recoveryshake.com 83 views

Introduction

In this problem, we are given an ellipse in the form of an equation and asked to identify the tangent lines to the ellipse. The equation of the ellipse is $9x^2 + 16y^2 - 144 = 0$ . To find the tangent lines, we need to first find the equation of the ellipse in standard form. Then, we can use the properties of the ellipse to find the tangent lines.

Standard Form of the Ellipse

To find the standard form of the ellipse, we need to complete the square for both the x and y terms.

Completing the Square for x

We start by completing the square for the x term. We can do this by adding and subtracting $(\frac{b}{2})^2$ inside the parentheses, where $b$ is the coefficient of the x term.

9x^2 + 16y^2 - 144 = 0
9(x^2 - \frac{144}{9}) + 16y^2 = 144
9(x^2 - 16) + 16y^2 = 144

Completing the Square for y

Next, we complete the square for the y term. We can do this by adding and subtracting $(\frac{c}{2})^2$ inside the parentheses, where $c$ is the coefficient of the y term.

9(x^2 - 16) + 16(y^2 - \frac{144}{16}) = 144
9(x^2 - 16) + 16(y^2 - 9) = 144

Simplifying the Equation

Now, we can simplify the equation by combining like terms.

9(x^2 - 16) + 16(y^2 - 9) = 144
9x^2 - 144 + 16y^2 - 144 = 144
9x^2 + 16y^2 = 432

Dividing by the Constant

To get the standard form of the ellipse, we need to divide both sides of the equation by the constant term.

\frac{9x^2}{432} + \frac{16y^2}{432} = \frac{432}{432}
\frac{x^2}{48} + \frac{y^2}{27} = 1

Properties of the Ellipse

Now that we have the standard form of the ellipse, we can use the properties of the ellipse to find the tangent lines.

Major and Minor Axes

The major axis of the ellipse is along the y-axis, and the minor axis is along the x-axis. The length of the major axis is $2\sqrt{27} = 6\sqrt{3}$ , and the length of the minor axis is $2\sqrt{48} = 8\sqrt{3}$ .

Foci

The foci of the ellipse are located at $(0, \pm c)$ , where $c = \sqrt{a^2 - b^2}$ . In this case, $a = \sqrt{27}$ and $b = \sqrt{48}$ , so $c = \sqrt{27 - 48} = \sqrt{-21}$ .

Tangent Lines

A tangent line to the ellipse is a line that intersects the ellipse at exactly one point. To find the tangent lines, we need to find the equations of the lines that are tangent to the ellipse.

Finding the Tangent Lines

To find the tangent lines, we can use the fact that the slope of the tangent line is equal to the slope of the radius at the point of tangency. We can find the slope of the radius by finding the derivative of the equation of the ellipse.

Derivative of the Equation of the Ellipse

To find the derivative of the equation of the ellipse, we can use implicit differentiation.

\frac{d}{dx}(\frac{x^2}{48} + \frac{y^2}{27}) = \frac{d}{dx}(1)
\frac{1}{48}x + \frac{1}{27}y\frac{dy}{dx} = 0
\frac{dy}{dx} = -\frac{27}{48}x

Finding the Slope of the Tangent Line

Now that we have the derivative of the equation of the ellipse, we can find the slope of the tangent line by finding the slope of the radius at the point of tangency.

m = -\frac{27}{48}x

Finding the Equation of the Tangent Line

Now that we have the slope of the tangent line, we can find the equation of the tangent line by using the point-slope form of a line.

y - y_1 = m(x - x_1)
y - y_1 = -\frac{27}{48}x(x - x_1)

Substituting the Values

We can substitute the values of $x_1$ and $y_1$ into the equation of the tangent line to find the equation of the tangent line.

y - y_1 = -\frac{27}{48}x(x - x_1)
y - 0 = -\frac{27}{48}x(x - 0)
y = -\frac{27}{48}x^2