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We are given the equation of an ellipse:

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

and asked to find the equations of the tangent and normal lines to this ellipse through the point \((10, 3)\).

### Step 1: Equation of the tangent line
The general formula for the tangent line to an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) at a point \((x_1, y_1)\) on the ellipse is:

\[
\frac{x_1 x}{a^2} + \frac{y_1 y}{b^2} = 1
\]

Here, the given ellipse is \(\frac{x^2}{81} + \frac{y^2}{9} = 1\), which means \(a^2 = 81\) and \(b^2 = 9\), so \(a = 9\) and \(b = 3\).

Substituting \(x_1 = 10\) and \(y_1 = 3\) into the formula for the tangent line, we get:

\[
\frac{10x}{81} + \frac{3y}{9} = 1
\]

Simplifying:

\[
\frac{10x}{81} + \frac{y}{3} = 1
\]

Multiplying through by 81 to eliminate the denominators:

\[
10x + 27y = 81
\]

Thus, the equation of the tangent line is:

\[
10x + 27y = 81
\]

### Step 2: Equation of the normal line
The slope of the tangent line is the negative reciprocal of the normal line. First, we need to find the slope of the tangent line.

Rewriting the tangent equation \(10x + 27y = 81\) in slope-intercept form:

\[
27y = -10x + 81
\]

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

Thus, the slope of the tangent line is \(-\frac{10}{27}\), so the slope of the normal line will be the negative reciprocal, which is \(\frac{27}{10}\).

Now, using the point-slope form of the line equation, the equation of the normal line passing through \((10, 3)\) is:

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

Simplifying:

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

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

Thus, the equation of the normal line is:

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

### Final answer:
- The equation of the tangent line is: \(10x + 27y = 81\)
- The equation of the normal line is: \(y = \frac{27}{10}x - 24\)

Would you like further explanation on any of the steps?

Here are 5 related questions for deeper exploration:
1. How would the tangent line equation change if the point was not on the ellipse?
2. What is the general formula for a normal line to an ellipse?
3. How can we geometrically interpret the relationship between the tangent and normal lines?
4. What would the tangent and normal lines look like if the ellipse equation had different coefficients?
5. How do the slopes of the tangent and normal lines vary as the point of tangency moves around the ellipse?

**Tip:** When finding the equation of a tangent to an ellipse, always ensure the point lies on the ellipse by checking if it satisfies the ellipse equation!

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 tangent and normal lines to the ellipse x^2 + 9y^2 = 81 at the point (10, 3). Detailed solution involving key geometry and calculus concepts such as the tangent line formula and the slope of the normal line.

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