If the normal to the rectangular hyperbola `xy = 4` at the point `(2t, (2)/(t_(1)))` meets the curve again at `(2t_(2), (2)/(t_(2)))`, then

Options

A. `t_(1)""^(3)t_(2) =1`
B. `t_(1)""^(3)t_(2) =-1`
C. `t_(2)""^(3)t_(1) =1`
D. `t_(1)t_(2)^(3) = -1`

Answer

To solve the problem step-by-step, we will follow the approach outlined in the video transcript.

### Step 1: Understand the given hyperbola
The equation of the rectangular hyperbola is given as:
\[ xy = 4 \]
We need to find the normal to this hyperbola at the point \( (2t_1, \frac{2}{t_1}) \).

### Step 2: Find the slope of the tangent
To find the slope of the tangent to the hyperbola, we differentiate the equation implicitly:
\[ \frac{d}{dx}(xy) = \frac{d}{dx}(4) \]
Using the product rule:
\[ x \frac{dy}{dx} + y = 0 \]
Thus, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = -\frac{y}{x} \]

### Step 3: Substitute the point into the derivative
Substituting the point \( (2t_1, \frac{2}{t_1}) \) into the derivative:
\[ \frac{dy}{dx} = -\frac{\frac{2}{t_1}}{2t_1} = -\frac{1}{t_1^2} \]
This gives us the slope of the tangent line at that point.

### Step 4: Find the slope of the normal
The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\left(-\frac{1}{t_1^2}\right)} = t_1^2 \]

### Step 5: Write the equation of the normal
Using the point-slope form of the line, the equation of the normal at the point \( (2t_1, \frac{2}{t_1}) \) is:
\[ y - \frac{2}{t_1} = t_1^2 (x - 2t_1) \]

### Step 6: Rearranging the normal equation
Rearranging gives:
\[ y - \frac{2}{t_1} = t_1^2 x - 2t_1^3 \]
Thus,
\[ y = t_1^2 x - 2t_1^3 + \frac{2}{t_1} \]

### Step 7: Substitute the second point into the normal equation
We know the normal meets the hyperbola again at the point \( (2t_2, \frac{2}{t_2}) \). Substitute \( x = 2t_2 \) and \( y = \frac{2}{t_2} \) into the normal equation:
\[ \frac{2}{t_2} = t_1^2 (2t_2) - 2t_1^3 + \frac{2}{t_1} \]

### Step 8: Simplifying the equation
Rearranging gives:
\[ \frac{2}{t_2} - 2t_1^2 t_2 + 2t_1^3 - \frac{2}{t_1} = 0 \]
Multiplying through by \( t_1 t_2 \) to eliminate the fractions:
\[ 2t_1 - 2t_1^2 t_2^2 + 2t_1^4 t_2 - 2t_2 = 0 \]

### Step 9: Factor the equation
Rearranging gives:
\[ 2t_1^4 t_2 - 2t_1^2 t_2^2 + 2t_1 - 2t_2 = 0 \]
Factoring out common terms:
\[ 2(t_1^4 t_2 - t_1^2 t_2^2 + t_1 - t_2) = 0 \]

### Step 10: Solve for \( t_1 \) and \( t_2 \)
This leads to the condition:
\[ t_1^3 t_2 = -1 \]

### Final Answer
Thus, the final condition we derive is:
\[ t_1^3 t_2 = -1 \]