Three normals are drawn from the point `(14,7)` to the curve `y^2-16x-8y=0`. Find the coordinates of the feet of the normals.

Options

A. (3,-4)
B. (8,16)
C. (0,0)
D. (2,2)

Answer

To solve the problem of finding the coordinates of the feet of the normals drawn from the point (14, 7) to the curve given by the equation \( y^2 - 16x - 8y = 0 \), we will follow these steps:

### Step 1: Rewrite the Parabola Equation
We start by rewriting the equation of the parabola in a standard form. The given equation is:
\[ y^2 - 8y - 16x = 0 \]
We can rearrange this to:
\[ y^2 - 8y = 16x \]
Next, we complete the square for the \(y\) terms:
\[ (y - 4)^2 - 16 = 16x \]
This simplifies to:
\[ (y - 4)^2 = 16(x + 1) \]
This shows that the parabola opens to the right with vertex at \((-1, 4)\).

### Step 2: Equation of the Normal
The equation of the normal to the parabola \(y - k = a(x - h)\) can be expressed as:
\[ y - 4 = m(x + 1) - 2am \]
where \(a = 4\) (from the standard form of the parabola) and \(m\) is the slope of the normal.

### Step 3: Substitute the Point (14, 7)
Since the normal passes through the point (14, 7), we substitute these coordinates into the normal equation:
\[ 7 - 4 = m(14 + 1) - 8m \]
This simplifies to:
\[ 3 = 15m - 8m \]
\[ 3 = 7m \]
Thus, we find:
\[ m = \frac{3}{7} \]

### Step 4: Form the Cubic Equation
To find the slopes of the normals, we need to derive a cubic equation. The general form of the cubic equation in \(m\) is:
\[ 4m^3 - 7m + 3 = 0 \]
We can use the fact that \(m = 1\) is one root (as shown in the video). We can factor the cubic polynomial:
\[ 4m^3 - 7m + 3 = (m - 1)(4m^2 + 4m - 3) \]

### Step 5: Solve the Quadratic for Remaining Roots
Now we solve the quadratic \(4m^2 + 4m - 3 = 0\) using the quadratic formula:
\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 48}}{8} = \frac{-4 \pm 8}{8} \]
This gives us:
\[ m_2 = \frac{4}{8} = \frac{1}{2}, \quad m_3 = \frac{-12}{8} = -\frac{3}{2} \]

### Step 6: Find Coordinates of Feet of Normals
Now we find the coordinates of the feet of the normals for each slope \(m\).

1. **For \(m_1 = 1\)**:
\[ x = -1 + 4(1^2) = 3, \quad y = 4 - 8 = -4 \quad \Rightarrow \quad P(3, -4) \]

2. **For \(m_2 = \frac{1}{2}\)**:
\[ x = -1 + 4\left(\frac{1}{2}\right)^2 = 0, \quad y = 4 - 4 = 0 \quad \Rightarrow \quad Q(0, 0) \]

3. **For \(m_3 = -\frac{3}{2}\)**:
\[ x = -1 + 4\left(-\frac{3}{2}\right)^2 = 8, \quad y = 4 + 12 = 16 \quad \Rightarrow \quad R(8, 16) \]

### Final Coordinates of the Feet of Normals
Thus, the coordinates of the feet of the normals are:
- \(P(3, -4)\)
- \(Q(0, 0)\)
- \(R(8, 16)\)