determine the equation of the tangent of the elipse 40xsquared+9ysquared =40 with a slope of minus 2/9

We are given the equation of an ellipse:

\[
40x^2 + 9y^2 = 40
\]

We need to determine the equation of the tangent line to this ellipse that has a slope of \( m = -\frac{2}{9} \).

### Step 1: Standard Form of the Ellipse
To begin, we simplify the given equation of the ellipse by dividing through by 40:

\[
\frac{x^2}{1} + \frac{y^2}{\frac{40}{9}} = 1
\]

This is the equation of an ellipse in standard form:

\[
\frac{x^2}{1} + \frac{y^2}{\frac{40}{9}} = 1
\]

where the semi-major axis is along the \( y \)-axis and the semi-minor axis is along the \( x \)-axis.

### Step 2: Equation of the Tangent Line
The general equation of a tangent line to the ellipse at a point \( (x_1, y_1) \) is given by:

\[
\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1
\]

where \( a^2 = 1 \) and \( b^2 = \frac{40}{9} \). Substituting these values, the equation of the tangent line becomes:

\[
x x_1 + \frac{9}{40} y y_1 = 1
\]

### Step 3: Condition for the Slope
We know that the slope of the tangent line is \( -\frac{2}{9} \), so the equation of the tangent line is of the form:

\[
y = -\frac{2}{9}x + c
\]

We now need to find the appropriate point \( (x_1, y_1) \) on the ellipse where the tangent has this slope. To do so, we will use implicit differentiation of the ellipse equation.

### Step 4: Implicit Differentiation
We differentiate the ellipse equation \( 40x^2 + 9y^2 = 40 \) with respect to \( x \):

\[
80x + 18y \frac{dy}{dx} = 0
\]

Solving for \( \frac{dy}{dx} \), the slope of the tangent line is:

\[
\frac{dy}{dx} = -\frac{80x}{18y} = -\frac{40x}{9y}
\]

We are given that the slope of the tangent line is \( -\frac{2}{9} \), so we set:

\[
-\frac{40x}{9y} = -\frac{2}{9}
\]

Simplifying:

\[
40x = 2y
\]

\[
y = 20x
\]

### Step 5: Substituting Back into the Ellipse Equation
Now substitute \( y = 20x \) into the ellipse equation:

\[
40x^2 + 9(20x)^2 = 40
\]

\[
40x^2 + 9(400x^2) = 40
\]

\[
40x^2 + 3600x^2 = 40
\]

\[
3640x^2 = 40
\]

\[
x^2 = \frac{40}{3640}
\]

\[
x^2 = \frac{1}{91}
\]

\[
x = \pm \frac{1}{\sqrt{91}}
\]

### Step 6: Finding \( y_1 \)
Now, using \( y = 20x \), we find \( y_1 \):

\[
y = 20 \times \frac{1}{\sqrt{91}} = \frac{20}{\sqrt{91}}
\]

Thus, the points of tangency are \( \left( \frac{1}{\sqrt{91}}, \frac{20}{\sqrt{91}} \right) \) and \( \left( -\frac{1}{\sqrt{91}}, -\frac{20}{\sqrt{91}} \right) \).

### Step 7: Equation of the Tangent Line
Finally, the equation of the tangent line is:

\[
y - y_1 = m(x - x_1)
\]

Substitute \( m = -\frac{2}{9} \) and the point \( \left( \frac{1}{\sqrt{91}}, \frac{20}{\sqrt{91}} \right) \):

\[
y - \frac{20}{\sqrt{91}} = -\frac{2}{9} \left( x - \frac{1}{\sqrt{91}} \right)
\]

This is the equation of the tangent line.

Would you like further details or clarification?

Here are five related questions to expand on this concept:
1. How do you determine the equation of a tangent line to any general conic section?
2. What are the applications of tangents in geometry and physics?
3. How does implicit differentiation help in finding tangents to curves?
4. Can you explain the significance of the slope in geometric contexts?
5. How can the method used here be applied to ellipses in different orientations?

**Tip**: Implicit differentiation is a powerful tool for finding slopes of tangents to implicitly defined curves.

Determine the equation of the tangent of the ellipse 40x^2 + 9y^2 = 40 with a slope of -2/9.

Learn how to find the equation of the tangent line to the ellipse 40x^2 + 9y^2 = 40 with a slope of -2/9 using implicit differentiation and conic section formulas. A detailed step-by-step solution explains the process. Suitable for advanced high school students in Grades 11-12.

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