Find the equation to the tangent and normal
at the point `(1, (4)/(3))` of the ellipse ` 4 x^(2) + 9y^(2) = 20`

Answer

To find the equations of the tangent and normal at the point \((1, \frac{4}{3})\) of the ellipse given by \(4x^2 + 9y^2 = 20\), we can follow these steps:

### Step 1: Rewrite the equation of the ellipse in standard form
We start with the equation of the ellipse:
\[ 4x^2 + 9y^2 = 20 \]
To convert it to standard form, divide both sides by 20:
\[ \frac{x^2}{5} + \frac{y^2}{\frac{20}{9}} = 1 \]
This gives us the standard form of the ellipse:
\[ \frac{x^2}{5} + \frac{y^2}{\frac{20}{9}} = 1 \]
Here, \(a^2 = 5\) and \(b^2 = \frac{20}{9}\).

### Step 2: Identify the coordinates of the point on the ellipse
The point given is \((1, \frac{4}{3})\). We will use this point to find the equations of the tangent and normal.

### Step 3: Use the formula for the tangent line at a point on the ellipse
The general equation of the tangent line to the ellipse at the point \((x_1, y_1)\) is given by:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
Substituting \(x_1 = 1\), \(y_1 = \frac{4}{3}\), \(a^2 = 5\), and \(b^2 = \frac{20}{9}\):
\[ \frac{x \cdot 1}{5} + \frac{y \cdot \frac{4}{3}}{\frac{20}{9}} = 1 \]
This simplifies to:
\[ \frac{x}{5} + \frac{4y}{3} \cdot \frac{9}{20} = 1 \]
\[ \frac{x}{5} + \frac{3y}{5} = 1 \]
Multiplying through by 5 to eliminate the denominators:
\[ x + 3y = 5 \]
Thus, the equation of the tangent line is:
\[ \boxed{x + 3y = 5} \]

### Step 4: Use the formula for the normal line at a point on the ellipse
The general equation of the normal line to the ellipse at the point \((x_1, y_1)\) is given by:
\[ a^2 \frac{x}{x_1} - b^2 \frac{y}{y_1} = a^2 - b^2 \]
Substituting \(x_1 = 1\), \(y_1 = \frac{4}{3}\), \(a^2 = 5\), and \(b^2 = \frac{20}{9}\):
\[ 5 \cdot \frac{x}{1} - \frac{20}{9} \cdot \frac{y}{\frac{4}{3}} = 5 - \frac{20}{9} \]
This simplifies to:
\[ 5x - \frac{20y}{12} = 5 - \frac{20}{9} \]
\[ 5x - \frac{5y}{3} = 5 - \frac{20}{9} \]
To simplify \(5 - \frac{20}{9}\), we convert 5 to a fraction:
\[ 5 = \frac{45}{9} \Rightarrow \frac{45}{9} - \frac{20}{9} = \frac{25}{9} \]
Thus, we have:
\[ 5x - \frac{5y}{3} = \frac{25}{9} \]
Multiplying through by 9 to eliminate the fraction:
\[ 45x - 15y = 25 \]
Rearranging gives us:
\[ 15y = 45x - 25 \]
Dividing through by 5:
\[ 3y = 9x - 5 \]
Thus, the equation of the normal line is:
\[ \boxed{3y = 9x - 5} \]