Distance between the parallel tangents having slope `-(4)/(3)` to the ellipse `(x^(2))/(18)+(y^(2))/(32) = 1` , is

Options

A. `24`
B. `(24)/(5)`
C. `(48)/(5)`
D. `(96)/(5)`

Explanation

To find the distance between the parallel tangents having slope \(-\frac{4}{3}\) to the ellipse given by the equation

\[ \frac{x^2}{18} + \frac{y^2}{32} = 1, \]

we can follow these steps:

### Step 1: Identify the parameters of the ellipse
The equation of the ellipse can be written in the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where:
- \(a^2 = 18\) (thus \(a = \sqrt{18} = 3\sqrt{2}\))
- \(b^2 = 32\) (thus \(b = \sqrt{32} = 4\sqrt{2}\))

### Step 2: Write the equation of the tangent
The equation of the tangent to the ellipse at a slope \(m\) is given by:

\[ y = mx \pm \sqrt{a^2 m^2 + b^2}. \]

### Step 3: Substitute the values
Here, \(m = -\frac{4}{3}\), \(a^2 = 18\), and \(b^2 = 32\). We substitute these values into the tangent equation:

\[ y = -\frac{4}{3}x \pm \sqrt{18 \left(-\frac{4}{3}\right)^2 + 32}. \]

Calculating \(m^2\):

\[ m^2 = \left(-\frac{4}{3}\right)^2 = \frac{16}{9}. \]

Now substituting \(m^2\) into the equation:

\[ y = -\frac{4}{3}x \pm \sqrt{18 \cdot \frac{16}{9} + 32}. \]

### Step 4: Simplify the expression under the square root
Calculating \(18 \cdot \frac{16}{9}\):

\[ 18 \cdot \frac{16}{9} = 2 \cdot 16 = 32. \]

Now, adding \(32\):

\[ 32 + 32 = 64. \]

So, we have:

\[ y = -\frac{4}{3}x \pm \sqrt{64} = -\frac{4}{3}x \pm 8. \]

### Step 5: Write the equations of the tangents
Thus, the two tangent lines are:

1. \(y = -\frac{4}{3}x + 8\)
2. \(y = -\frac{4}{3}x - 8\)

### Step 6: Calculate the distance between the two tangents
The distance \(d\) between two parallel lines of the form \(y = mx + c_1\) and \(y = mx + c_2\) is given by the formula:

\[ d = \frac{|c_1 - c_2|}{\sqrt{1 + m^2}}. \]

Here, \(c_1 = 8\) and \(c_2 = -8\):

\[ d = \frac{|8 - (-8)|}{\sqrt{1 + \left(-\frac{4}{3}\right)^2}} = \frac{|8 + 8|}{\sqrt{1 + \frac{16}{9}}} = \frac{16}{\sqrt{\frac{25}{9}}}. \]

### Step 7: Simplify the distance
Calculating the denominator:

\[ \sqrt{\frac{25}{9}} = \frac{5}{3}. \]

So, we have:

\[ d = \frac{16}{\frac{5}{3}} = 16 \cdot \frac{3}{5} = \frac{48}{5}. \]

### Final Answer
The distance between the parallel tangents is

\[ \frac{48}{5}. \]

---