a. Determine an equation of the tangent line and the normal line at the given point on the curve.
b. Graph the tangent and normal lines on the given graph.
11 left parenthesis x squared plus y squared right parenthesis squared equals 3721 left parenthesis x squared minus y squared right parenthesis11x2+y22=3721x2−y2​; left parenthesis 6 comma 5 right parenthesis(6,5)

Let's go through the steps required to solve the problem:

### Part a: **Determine the equation of the tangent line and the normal line at the given point (6, 5) on the curve.**

The given equation is:
\[
\left( x^2 + y^2 \right)^2 = 3721 \left( x^2 - y^2 \right)
\]

We'll follow these steps:
1. **Implicit Differentiation**: Differentiate both sides of the equation with respect to \(x\) to find the slope of the tangent line.
2. **Find the Slope of the Tangent Line**: Use the point (6, 5) to calculate the slope.
3. **Equation of the Tangent Line**: Use the point-slope form of the line equation.
4. **Equation of the Normal Line**: The normal line is perpendicular to the tangent line. If the slope of the tangent line is \(m\), the slope of the normal line is \(-\frac{1}{m}\).

#### Step 1: Implicit Differentiation

Differentiate both sides of the equation with respect to \(x\).

The left-hand side is:
\[
\frac{d}{dx} \left( x^2 + y^2 \right)^2 = 2 \left( x^2 + y^2 \right) \cdot \left( 2x + 2y \frac{dy}{dx} \right)
\]
The right-hand side is:
\[
\frac{d}{dx} \left( 3721 \left( x^2 - y^2 \right) \right) = 3721 \left( 2x - 2y \frac{dy}{dx} \right)
\]

This results in:
\[
2 \left( x^2 + y^2 \right) \cdot \left( 2x + 2y \frac{dy}{dx} \right) = 3721 \left( 2x - 2y \frac{dy}{dx} \right)
\]

Simplify:
\[
4 \left( x^2 + y^2 \right) \cdot \left( x + y \frac{dy}{dx} \right) = 7442 \left( x - y \frac{dy}{dx} \right)
\]

#### Step 2: Find the Slope at \( (6, 5) \)

Now, substitute \( x = 6 \) and \( y = 5 \) into the differentiated equation. This will allow us to find \( \frac{dy}{dx} \), which is the slope of the tangent line at the point (6, 5).

Substitute \( x = 6 \) and \( y = 5 \) into the original equation to find the values of \( x^2 + y^2 \):
\[
(6^2 + 5^2)^2 = 3721 \cdot (6^2 - 5^2)
\]
Simplifying this will allow us to calculate the slope of the tangent line.

Let me compute this for you.The slope of the tangent line at the point \( (6, 5) \) is \( \frac{118}{105} \).

#### Step 3: Equation of the Tangent Line
We can now use the point-slope form of the equation of a line:

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

Substituting \( m = \frac{118}{105} \) and the point \( (6, 5) \):

\[
y - 5 = \frac{118}{105}(x - 6)
\]

This is the equation of the tangent line.

#### Step 4: Equation of the Normal Line
The slope of the normal line is the negative reciprocal of the tangent slope:

\[
m_{\text{normal}} = -\frac{1}{\frac{118}{105}} = -\frac{105}{118}
\]

Using the point-slope form again for the normal line:

\[
y - 5 = -\frac{105}{118}(x - 6)
\]

This is the equation of the normal line.

---

### Part b: **Graph the Tangent and Normal Lines**

Would you like me to plot the graph of the curve along with the tangent and normal lines at the point \( (6, 5) \)?

Determine an equation of the tangent line and the normal line at the given point on the curve. Graph the tangent and normal lines on the given graph. 11(x^2 + y^2)^2 = 3721(x^2 - y^2); (6,5)

Learn how to find the equations of the tangent and normal lines at the point (6,5) on the curve (x^2 + y^2)^2 = 3721(x^2 - y^2). The solution involves implicit differentiation and the use of point-slope form for both lines, suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation