Find the equation of tangent and normal to the ellipse `x^2+8y^2=33` at (-1,2).

Answer

To find the equation of the tangent and normal to the ellipse \(x^2 + 8y^2 = 33\) at the point \((-1, 2)\), we will follow these steps:

### Step 1: Verify the point lies on the ellipse
First, we need to check if the point \((-1, 2)\) lies on the ellipse.

Substituting \(x = -1\) and \(y = 2\) into the ellipse equation:
\[ (-1)^2 + 8(2)^2 = 1 + 8 \cdot 4 = 1 + 32 = 33 \]
Since this is true, the point \((-1, 2)\) lies on the ellipse.

### Step 2: Differentiate the ellipse equation
Next, we differentiate the equation of the ellipse implicitly to find the slope of the tangent line.

The equation is:
\[ x^2 + 8y^2 = 33 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(8y^2) = \frac{d}{dx}(33) \]
This gives:
\[ 2x + 16y \frac{dy}{dx} = 0 \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to solve for \(\frac{dy}{dx}\):
\[ 16y \frac{dy}{dx} = -2x \]
\[ \frac{dy}{dx} = -\frac{2x}{16y} = -\frac{x}{8y} \]

### Step 4: Find the slope at the given point
Now we substitute the point \((-1, 2)\) into the derivative to find the slope of the tangent line:
\[ \frac{dy}{dx} = -\frac{-1}{8 \cdot 2} = \frac{1}{16} \]
Thus, the slope \(m_t\) of the tangent line at the point \((-1, 2)\) is \(\frac{1}{16}\).

### Step 5: Write the equation of the tangent line
Using the point-slope form of the equation of a line:
\[ y - y_1 = m_t(x - x_1) \]
Substituting \(m_t = \frac{1}{16}\), \(x_1 = -1\), and \(y_1 = 2\):
\[ y - 2 = \frac{1}{16}(x + 1) \]
Multiplying through by 16 to eliminate the fraction:
\[ 16(y - 2) = x + 1 \]
This simplifies to:
\[ 16y - 32 = x + 1 \]
Rearranging gives:
\[ x - 16y + 33 = 0 \]
Thus, the equation of the tangent line is:
\[ \boxed{x - 16y + 33 = 0} \]

### Step 6: Find the slope of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
\[ m_n = -\frac{1}{m_t} = -16 \]

### Step 7: Write the equation of the normal line
Using the point-slope form again for the normal line:
\[ y - y_1 = m_n(x - x_1) \]
Substituting \(m_n = -16\), \(x_1 = -1\), and \(y_1 = 2\):
\[ y - 2 = -16(x + 1) \]
Distributing gives:
\[ y - 2 = -16x - 16 \]
Rearranging gives:
\[ 16x + y + 14 = 0 \]
Thus, the equation of the normal line is:
\[ \boxed{16x + y + 14 = 0} \]