The line `4sqrt(2x)-5y=40` touches the hyperbola `(x^(2))/(100)-(y^(2)_)/(64)` =1 at the point

Options

A. `10,8sqrt(2)`
B. `10,8sqrt(2)`
C. `20,8sqrt(2)`
D. `(20)sqrt(3),(8)/sqrt(3)`

Answer

To solve the problem, we need to find the point at which the line \( 4\sqrt{2}x - 5y = 40 \) touches the hyperbola \( \frac{x^2}{100} - \frac{y^2}{64} = 1 \).

### Step 1: Identify the parameters of the hyperbola
The hyperbola is given in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
From the equation, we can identify:
- \( a^2 = 100 \) which gives \( a = 10 \)
- \( b^2 = 64 \) which gives \( b = 8 \)

### Step 2: Write the equation of the tangent to the hyperbola
The equation of the tangent to the hyperbola at a point \( (x_1, y_1) \) is given by:
\[ \frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1 \]
Substituting \( a^2 \) and \( b^2 \):
\[ \frac{x x_1}{100} - \frac{y y_1}{64} = 1 \]

### Step 3: Rewrite the line equation
The line equation can be rewritten in slope-intercept form:
\[ 5y = 4\sqrt{2}x - 40 \implies y = \frac{4\sqrt{2}}{5}x - 8 \]
The slope of the line is \( \frac{4\sqrt{2}}{5} \).

### Step 4: Find the slope of the tangent line
The slope of the tangent line at the point \( (x_1, y_1) \) on the hyperbola can also be expressed in terms of \( \theta \):
\[ \text{slope} = \frac{b}{a} \cdot \frac{dx}{dy} = \frac{8 \sec \theta}{10 \tan \theta} \]
This simplifies to:
\[ \frac{4}{5} \cdot \frac{\sec \theta}{\tan \theta} = \frac{4}{5} \cdot \frac{1}{\sin \theta} \]

### Step 5: Set the slopes equal
Equating the slopes from the line and the hyperbola gives us:
\[ \frac{4\sqrt{2}}{5} = \frac{4}{5} \cdot \frac{1}{\sin \theta} \]
This implies:
\[ \sin \theta = \sqrt{2} \]
Since \( \sin \theta \) cannot exceed 1, we need to find \( \theta \) such that:
\[ \cos \theta = \frac{1}{\sqrt{2}} \implies \theta = 45^\circ \]

### Step 6: Find the coordinates of the point of tangency
Using \( \theta = 45^\circ \):
\[ x_1 = a \sec \theta = 10 \cdot \sqrt{2} = 10\sqrt{2} \]
\[ y_1 = b \tan \theta = 8 \cdot 1 = 8 \]

### Final Answer
The point at which the line touches the hyperbola is:
\[ (10\sqrt{2}, 8) \]