The normal at any point `P(x_1,y_1)` of curve is a line perpendicular to tangent at the point `P(x_1,y_1)`. In case of rectangular hyperbola `xy=c^2`, the equation of normal at `(ct,(c )/(t))` is `xt^3-yt-ct^4+c=0`. The shortest distance between any two curves always along the common normal.
If normal at (5, 3) of rectangular hyperbola `xy-y-2x-2=0` intersect it again at a point:

Options

A. `(-1,0)`
B. `(-1,1)`
C. `(0,-2)`
D. `((3)/(4),-14)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Find the equation of the hyperbola
The given hyperbola is \( xy - y - 2x - 2 = 0 \). We can rearrange this to express \( y \) in terms of \( x \):
\[ y = \frac{2x + 2}{x - 1} \]

### Step 2: Differentiate to find the slope of the tangent
We will differentiate \( y \) with respect to \( x \) to find the slope of the tangent line at the point \( (5, 3) \).

Using the quotient rule:
\[ y' = \frac{(x - 1)(2) - (2x + 2)(1)}{(x - 1)^2} \]
Simplifying this:
\[ y' = \frac{2x - 2 - 2x - 2}{(x - 1)^2} = \frac{-4}{(x - 1)^2} \]

### Step 3: Find the slope of the tangent at the point (5, 3)
Substituting \( x = 5 \):
\[ y' = \frac{-4}{(5 - 1)^2} = \frac{-4}{16} = -\frac{1}{4} \]
Thus, the slope of the tangent at the point \( (5, 3) \) is \( -\frac{1}{4} \).

### Step 4: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ m_{\text{normal}} = -\frac{1}{(-\frac{1}{4})} = 4 \]

### Step 5: Write the equation of the normal line
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (5, 3) \) and \( m = 4 \):
\[ y - 3 = 4(x - 5) \]
This simplifies to:
\[ y = 4x - 20 + 3 \implies y = 4x - 17 \]

### Step 6: Find the intersection of the normal with the hyperbola
We substitute \( y = 4x - 17 \) into the hyperbola equation:
\[ 4x - 17 = \frac{2x + 2}{x - 1} \]
Cross-multiplying:
\[ (4x - 17)(x - 1) = 2x + 2 \]
Expanding and rearranging:
\[ 4x^2 - 4x - 17x + 17 = 2x + 2 \implies 4x^2 - 23x + 15 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{23 \pm \sqrt{(-23)^2 - 4 \cdot 4 \cdot 15}}{2 \cdot 4} \]
Calculating the discriminant:
\[ 529 - 240 = 289 \implies \sqrt{289} = 17 \]
Thus:
\[ x = \frac{23 \pm 17}{8} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{40}{8} = 5 \)
2. \( x = \frac{6}{8} = \frac{3}{4} \)

### Step 8: Find the corresponding \( y \) value for \( x = \frac{3}{4} \)
Substituting \( x = \frac{3}{4} \) into the normal equation:
\[ y = 4\left(\frac{3}{4}\right) - 17 = 3 - 17 = -14 \]

### Final Result
The point where the normal intersects the hyperbola again is:
\[ \left(\frac{3}{4}, -14\right) \]