Find the equation of the tangent to the curve `y=6-x^(2)`, where the normal is parallel to the line `x-4y+3=0`.

Answer

To find the equation of the tangent to the curve \( y = 6 - x^2 \) where the normal is parallel to the line \( x - 4y + 3 = 0 \), we can follow these steps:

### Step 1: Find the slope of the normal line
The equation of the line is given as \( x - 4y + 3 = 0 \). We can rewrite this in slope-intercept form (i.e., \( y = mx + b \)) to find the slope.

\[ x - 4y + 3 = 0 \implies 4y = x + 3 \implies y = \frac{1}{4}x + \frac{3}{4} \]

Thus, the slope of the normal line \( m_1 \) is \( \frac{1}{4} \).

### Step 2: Find the slope of the tangent line
Since the normal and tangent lines are perpendicular to each other, their slopes satisfy the relationship:

\[ m_1 \cdot m_2 = -1 \]

Where \( m_2 \) is the slope of the tangent. Substituting \( m_1 = \frac{1}{4} \):

\[ \frac{1}{4} \cdot m_2 = -1 \implies m_2 = -4 \]

### Step 3: Differentiate the curve to find the slope
The curve is given by \( y = 6 - x^2 \). We differentiate this with respect to \( x \):

\[ \frac{dy}{dx} = -2x \]

### Step 4: Set the derivative equal to the slope of the tangent
We set the derivative equal to the slope of the tangent we found:

\[ -2x = -4 \]

Solving for \( x \):

\[ 2x = 4 \implies x = 2 \]

### Step 5: Find the corresponding \( y \) value
Now, we substitute \( x = 2 \) back into the original curve equation to find the corresponding \( y \):

\[ y = 6 - (2)^2 = 6 - 4 = 2 \]

Thus, the point of tangency is \( (2, 2) \).

### Step 6: Write the equation of the tangent line
Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (2, 2) \) and \( m = -4 \):

\[ y - 2 = -4(x - 2) \]

Expanding this:

\[ y - 2 = -4x + 8 \implies y = -4x + 10 \]

### Step 7: Rearranging to standard form
To express this in standard form \( Ax + By + C = 0 \):

\[ 4x + y - 10 = 0 \]

Thus, the equation of the tangent line is:

\[ 4x + y - 10 = 0 \]