. A straight line touches the rectangular hyperbola `9x^2-9y^2=8` and the parabola `y^2= 32x`. An equation of the line is

Options

A. `9x+3y-8=infty`
B. `9x-3y+8=0`
C. `9x+3y+8=0`
D. `9x-3y-8=0`

Answer

To find the equation of the straight line that touches the rectangular hyperbola \(9x^2 - 9y^2 = 8\) and the parabola \(y^2 = 32x\), we can follow these steps:

### Step 1: Rewrite the equations in standard form
The hyperbola can be rewritten as:
\[ \frac{x^2}{\frac{8}{9}} - \frac{y^2}{\frac{8}{9}} = 1 \]
This indicates that \(a^2 = \frac{8}{9}\) and \(b^2 = \frac{8}{9}\).

For the parabola, we can rewrite it in the standard form:
\[ y^2 = 4ax \quad \text{where } 4a = 32 \Rightarrow a = 8 \]

### Step 2: Write the general equations of the tangents
For the hyperbola, the equation of the tangent line in slope-intercept form is:
\[ y = mx \pm \sqrt{a^2 m^2 - b^2} \]
Substituting \(a^2\) and \(b^2\):
\[ y = mx \pm \sqrt{\frac{8}{9} m^2 - \frac{8}{9}} = mx \pm \frac{2\sqrt{2}}{3} \sqrt{m^2 - 1} \]

For the parabola, the equation of the tangent line is:
\[ y = mx + \frac{a}{m} = mx + \frac{8}{m} \]

### Step 3: Set the constant terms equal
Since both lines are tangents to their respective curves, we can equate the constant terms:
\[ \frac{2\sqrt{2}}{3} \sqrt{m^2 - 1} = \frac{8}{m} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{2\sqrt{2}}{3}\right)^2 (m^2 - 1) = \left(\frac{8}{m}\right)^2 \]
This simplifies to:
\[ \frac{8}{9}(m^2 - 1) = \frac{64}{m^2} \]

### Step 5: Clear the fractions
Multiplying through by \(9m^2\) gives:
\[ 8m^2(m^2 - 1) = 576 \]
Expanding this results in:
\[ 8m^4 - 8m^2 - 576 = 0 \]

### Step 6: Divide by 8
Dividing the entire equation by 8:
\[ m^4 - m^2 - 72 = 0 \]

### Step 7: Substitute \(u = m^2\)
Let \(u = m^2\):
\[ u^2 - u - 72 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula:
\[ u = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-72)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 288}}{2} = \frac{1 \pm 17}{2} \]
This gives:
\[ u = 9 \quad \text{or} \quad u = -8 \]
Since \(u = m^2\), we discard \(u = -8\) and take \(u = 9\), giving \(m^2 = 9\) or \(m = \pm 3\).

### Step 9: Find the equations of the tangents
Substituting \(m = 3\) into the tangent equation for the parabola:
\[ y = 3x + \frac{8}{3} \]
Rearranging gives:
\[ 3y - 9x - 8 = 0 \]

Substituting \(m = -3\):
\[ y = -3x - \frac{8}{3} \]
Rearranging gives:
\[ 3y + 9x + 8 = 0 \]

### Final Answer
The equations of the tangents are:
1. \(9x - 3y + 8 = 0\)
2. \(9x + 3y + 8 = 0\)

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