If a hyperbola has vertices `(±6, 0)` and P(10, 16) lies on it, then the equation of normal at P is (A) 2x + 5y = 100 (B) 2x + 5y = 10 (C) 2x -5y = 100 (D) 5x +2y = 100

Options

A. 2x + 5y = 100
B. 2x + 5y = 10
C. 2x – 5y = 100
D. 5x + 2y = 100

Explanation

To find the equation of the normal to the hyperbola at the point P(10, 16), we will follow these steps:

### Step 1: Identify the hyperbola's equation
Given the vertices of the hyperbola at (±6, 0), we know that:
- The distance from the center to the vertices is \( a = 6 \).
- The standard form of the hyperbola is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Hence, \( a^2 = 36 \).

### Step 2: Use the point P(10, 16) to find \( b^2 \)
Since point P lies on the hyperbola, we can substitute \( x = 10 \) and \( y = 16 \) into the hyperbola's equation:
\[ \frac{10^2}{36} - \frac{16^2}{b^2} = 1 \]
Calculating the left side:
\[ \frac{100}{36} - \frac{256}{b^2} = 1 \]
This simplifies to:
\[ \frac{100}{36} - 1 = \frac{256}{b^2} \]
\[ \frac{100 - 36}{36} = \frac{256}{b^2} \]
\[ \frac{64}{36} = \frac{256}{b^2} \]

### Step 3: Solve for \( b^2 \)
Cross-multiplying gives:
\[ 64b^2 = 256 \times 36 \]
Calculating \( 256 \times 36 \):
\[ 256 \times 36 = 9216 \]
Thus:
\[ 64b^2 = 9216 \implies b^2 = \frac{9216}{64} = 144 \]
So, \( b = 12 \).

### Step 4: Write the hyperbola's equation
Now we can write the equation of the hyperbola:
\[ \frac{x^2}{36} - \frac{y^2}{144} = 1 \]

### Step 5: Find the slope of the tangent line at P
To find the slope of the tangent line, we differentiate the hyperbola's equation implicitly:
\[ \frac{d}{dx}\left(\frac{x^2}{36} - \frac{y^2}{144}\right) = 0 \]
This gives:
\[ \frac{2x}{36} - \frac{2y}{144} \cdot \frac{dy}{dx} = 0 \]
Simplifying:
\[ \frac{x}{18} - \frac{y}{72} \cdot \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ \frac{dy}{dx} = \frac{72x}{18y} = \frac{4x}{y} \]

### Step 6: Calculate the slope at point P(10, 16)
Substituting \( x = 10 \) and \( y = 16 \):
\[ \frac{dy}{dx} = \frac{4 \cdot 10}{16} = \frac{40}{16} = \frac{5}{2} \]
Thus, the slope of the tangent line at P is \( \frac{5}{2} \).

### Step 7: Find the slope of the normal line
The slope of the normal line \( m_n \) is the negative reciprocal of the slope of the tangent line:
\[ m_n = -\frac{1}{\frac{5}{2}} = -\frac{2}{5} \]

### Step 8: Write the equation of the normal line
Using the point-slope form of the equation of a line:
\[ y - y_1 = m_n(x - x_1) \]
Substituting \( m_n = -\frac{2}{5} \) and the point \( P(10, 16) \):
\[ y - 16 = -\frac{2}{5}(x - 10) \]
Multiplying through by 5 to eliminate the fraction:
\[ 5(y - 16) = -2(x - 10) \]
Expanding:
\[ 5y - 80 = -2x + 20 \]
Rearranging gives:
\[ 2x + 5y = 100 \]

### Final Answer
The equation of the normal at point P(10, 16) is:
\[ \boxed{2x + 5y = 100} \]