Tangents are drawn to the hyperbola `(x^(2))/(9)-(y^(2))/(4)=1` parallel to the straight line `2x-y=1`. The points of contact of the tangent on the hyperbola are

Options

A. `((9)/(2sqrt(2)),(1)/(sqrt(2)))`
B. `(-(9)/(2sqrt(2)),-(1)/(sqrt(2)))`
C. `(3sqrt(3),-2sqrt(2))`
D. `(-3sqrt(3),2sqrt(2))`

Answer

To find the points of contact of the tangents drawn to the hyperbola \(\frac{x^2}{9} - \frac{y^2}{4} = 1\) that are parallel to the line \(2x - y = 1\), we can follow these steps:

### Step 1: Find the slope of the given line
The equation of the line is given as \(2x - y = 1\). We can rewrite it in slope-intercept form (i.e., \(y = mx + b\)):
\[ y = 2x - 1 \]
From this, we can see that the slope \(m\) of the line is \(2\).

### Step 2: Find the derivative of the hyperbola
To find the slope of the tangent to the hyperbola, we differentiate the hyperbola equation:
\[ \frac{x^2}{9} - \frac{y^2}{4} = 1 \]
Differentiating both sides with respect to \(x\):
\[ \frac{2x}{9} - \frac{2y}{4} \frac{dy}{dx} = 0 \]
This simplifies to:
\[ \frac{2x}{9} = \frac{2y}{4} \frac{dy}{dx} \]
Thus:
\[ \frac{dy}{dx} = \frac{4x}{9y} \]

### Step 3: Set the derivative equal to the slope of the line
Since the tangents are parallel to the line, we set the derivative equal to the slope of the line:
\[ \frac{4x}{9y} = 2 \]
Cross-multiplying gives:
\[ 4x = 18y \quad \Rightarrow \quad x = \frac{9y}{2} \]

### Step 4: Substitute \(x\) back into the hyperbola equation
Now we substitute \(x = \frac{9y}{2}\) back into the hyperbola equation:
\[ \frac{\left(\frac{9y}{2}\right)^2}{9} - \frac{y^2}{4} = 1 \]
Calculating \(\left(\frac{9y}{2}\right)^2\):
\[ \frac{81y^2}{4} \div 9 = \frac{9y^2}{4} \]
Thus, we have:
\[ \frac{9y^2}{4} - \frac{y^2}{4} = 1 \]
This simplifies to:
\[ \frac{8y^2}{4} = 1 \quad \Rightarrow \quad 2y^2 = 1 \quad \Rightarrow \quad y^2 = \frac{1}{2} \]

### Step 5: Solve for \(y\)
Taking the square root gives:
\[ y = \pm \frac{1}{\sqrt{2}} \]

### Step 6: Solve for \(x\) using \(y\)
Now we substitute \(y\) back into the equation \(x = \frac{9y}{2}\):
1. For \(y = \frac{1}{\sqrt{2}}\):
\[ x = \frac{9 \cdot \frac{1}{\sqrt{2}}}{2} = \frac{9}{2\sqrt{2}} = \frac{9\sqrt{2}}{4} \]
Thus, one point of contact is \(\left(\frac{9\sqrt{2}}{4}, \frac{1}{\sqrt{2}}\right)\).

2. For \(y = -\frac{1}{\sqrt{2}}\):
\[ x = \frac{9 \cdot -\frac{1}{\sqrt{2}}}{2} = -\frac{9}{2\sqrt{2}} = -\frac{9\sqrt{2}}{4} \]
Thus, the other point of contact is \(\left(-\frac{9\sqrt{2}}{4}, -\frac{1}{\sqrt{2}}\right)\).

### Final Points of Contact
The points of contact of the tangents on the hyperbola are:
1. \(\left(\frac{9\sqrt{2}}{4}, \frac{1}{\sqrt{2}}\right)\)
2. \(\left(-\frac{9\sqrt{2}}{4}, -\frac{1}{\sqrt{2}}\right)\)

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