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We are asked to find the equations of the tangent and normal lines to the ellipse given by the equation:

\[
\frac{x^2}{81} + \frac{y^2}{9} = 1
\]

through the point \( (10, 3) \).

### Step 1: Differentiate the equation of the ellipse
The equation of the ellipse can be written as:

\[
x^2/81 + y^2/9 = 1.
\]

To find the slope of the tangent line, we implicitly differentiate both sides with respect to \(x\):

\[
\frac{d}{dx} \left( \frac{x^2}{81} + \frac{y^2}{9} \right) = \frac{d}{dx}(1).
\]

This results in:

\[
\frac{2x}{81} + \frac{2y}{9} \cdot \frac{dy}{dx} = 0.
\]

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

\[
\frac{2y}{9} \cdot \frac{dy}{dx} = - \frac{2x}{81}
\]

\[
\frac{dy}{dx} = - \frac{2x}{81} \cdot \frac{9}{2y}
\]

\[
\frac{dy}{dx} = - \frac{x}{9y}.
\]

So, the slope of the tangent line is:

\[
m_{\text{tangent}} = - \frac{x}{9y}.
\]

### Step 2: Plug in the point \( (10, 3) \)
We now substitute \( x = 10 \) and \( y = 3 \) into the formula for the slope of the tangent:

\[
m_{\text{tangent}} = - \frac{10}{9 \times 3} = - \frac{10}{27}.
\]

### Step 3: Equation of the tangent line
The equation of the tangent line is given by the point-slope form of a line:

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

where \( (x_1, y_1) = (10, 3) \) and \( m = -\frac{10}{27} \). Substituting these values:

\[
y - 3 = - \frac{10}{27}(x - 10).
\]

Simplifying:

\[
y - 3 = - \frac{10}{27}x + \frac{100}{27},
\]

\[
y = - \frac{10}{27}x + \frac{100}{27} + 3,
\]

\[
y = - \frac{10}{27}x + \frac{100}{27} + \frac{81}{27},
\]

\[
y = - \frac{10}{27}x + \frac{181}{27}.
\]

Thus, the equation of the tangent line is:

\[
y = - \frac{10}{27}x + \frac{181}{27}.
\]

### Step 4: Equation of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line is \( - \frac{10}{27} \), the slope of the normal line is:

\[
m_{\text{normal}} = \frac{27}{10}.
\]

Using the point-slope form of a line again with the point \( (10, 3) \):

\[
y - 3 = \frac{27}{10}(x - 10),
\]

Simplifying:

\[
y - 3 = \frac{27}{10}x - 27,
\]

\[
y = \frac{27}{10}x - 27 + 3,
\]

\[
y = \frac{27}{10}x - 24.
\]

Thus, the equation of the normal line is:

\[
y = \frac{27}{10}x - 24.
\]

### Final Answer:
- The equation of the tangent line is \( y = - \frac{10}{27}x + \frac{181}{27} \).
- The equation of the normal line is \( y = \frac{27}{10}x - 24 \).

Would you like more details, or do you have any questions?

Here are 5 related questions for further practice:
1. How do you find the tangent line to a hyperbola at a given point?
2. What is the geometric interpretation of the normal line in terms of curves?
3. How can you find the tangent and normal lines for a general ellipse centered at \( (h, k) \)?
4. How would the equations of the tangent and normal lines change if the ellipse were rotated?
5. Can you derive the equation for the tangent line to a parabola?

**Tip:** The slope of the normal line is always the negative reciprocal of the slope of the tangent line for curves.

Find the equations of the tangent and normal lines to the ellipse x^2 + 9y^2 = 81 through the point (10, 3).

Learn how to find the equations of the tangent and normal lines to the ellipse x^2 + 9y^2 = 81 through the point (10, 3). This step-by-step solution covers implicit differentiation and key concepts of tangents and normals to ellipses. Suitable for students in Grades 11-12 and college-level mathematics.

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