Equations of those tangents to `4x^(2) -9y^(2) = 36` which are prependicular to the straight line `2y + 5x = 10`, are

Options

A. `5(y -3) = (x - sqrt(117//4))`
B. `5(y-2) = 2(x- sqrt18)`
C. `5(y+2) =2(x- sqrt18)`
D. none of these

Answer

To find the equations of the tangents to the hyperbola given by \(4x^{2} - 9y^{2} = 36\) that are perpendicular to the line \(2y + 5x = 10\), we can follow these steps:

### Step 1: Determine the slope of the given line
The equation of the line is given as:
\[ 2y + 5x = 10 \]
Rearranging it into slope-intercept form \(y = mx + c\):
\[ 2y = -5x + 10 \implies y = -\frac{5}{2}x + 5 \]
Thus, the slope \(m\) of the line is:
\[ m = -\frac{5}{2} \]

### Step 2: Find the slope of the tangent line
Since we are looking for tangents that are perpendicular to this line, the slope \(m_t\) of the tangent must satisfy:
\[ m_t \cdot m = -1 \]
Substituting the value of \(m\):
\[ m_t \cdot \left(-\frac{5}{2}\right) = -1 \implies m_t = \frac{2}{5} \]

### Step 3: Differentiate the hyperbola to find the slope of the tangent
The equation of the hyperbola is:
\[ 4x^{2} - 9y^{2} = 36 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(4x^{2}) - \frac{d}{dx}(9y^{2}) = \frac{d}{dx}(36) \]
This gives:
\[ 8x - 18y\frac{dy}{dx} = 0 \]
Rearranging for \(\frac{dy}{dx}\):
\[ 18y\frac{dy}{dx} = 8x \implies \frac{dy}{dx} = \frac{8x}{18y} = \frac{4x}{9y} \]

### Step 4: Set the derivative equal to the slope of the tangent
At the point of tangency \((x_1, y_1)\), we have:
\[ \frac{4x_1}{9y_1} = \frac{2}{5} \]
Cross-multiplying gives:
\[ 20x_1 = 18y_1 \implies 10x_1 = 9y_1 \implies y_1 = \frac{10}{9}x_1 \]

### Step 5: Substitute \(y_1\) back into the hyperbola equation
Substituting \(y_1\) into the hyperbola equation:
\[ 4x_1^{2} - 9\left(\frac{10}{9}x_1\right)^{2} = 36 \]
Simplifying gives:
\[ 4x_1^{2} - 9 \cdot \frac{100}{81}x_1^{2} = 36 \]
\[ 4x_1^{2} - \frac{900}{81}x_1^{2} = 36 \]
Finding a common denominator:
\[ \frac{324}{81}x_1^{2} - \frac{900}{81}x_1^{2} = 36 \]
\[ \frac{-576}{81}x_1^{2} = 36 \]
Multiplying both sides by \(-81\):
\[ 576x_1^{2} = -2916 \]
This leads to:
\[ x_1^{2} = -\frac{2916}{576} \]
Since \(x_1^{2}\) cannot be negative, there are no real solutions for \(x_1\).

### Conclusion
Since there are no real values for \(y_1\) and \(x_1\), we conclude that there are no tangents to the hyperbola \(4x^{2} - 9y^{2} = 36\) that are perpendicular to the line \(2y + 5x = 10\).