The points on the ellipse `(x^(2))/(2)+(y^(2))/(10)=1` from which perpendicular tangents can be drawn to the hyperbola `(x^(2))/(5)-(y^(2))/(1) =1` is/are

Options

A. `(sqrt((3)/(2)),sqrt((5)/(2)))`
B. `(sqrt((3)/(2)),-sqrt((5)/(2)))`
C. `(-sqrt((3)/(2)),sqrt((5)/(2)))`
D. `(sqrt((5)/(2)),sqrt((3)/(2)))`

Explanation

To solve the problem step by step, we need to find the points on the ellipse \(\frac{x^2}{2} + \frac{y^2}{10} = 1\) from which perpendicular tangents can be drawn to the hyperbola \(\frac{x^2}{5} - \frac{y^2}{1} = 1\).

### Step 1: Find the equation of the director circle of the hyperbola
The equation of the director circle for a hyperbola given in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) is given by:

\[ x^2 + y^2 = a^2 - b^2 \]

For the hyperbola \(\frac{x^2}{5} - \frac{y^2}{1} = 1\):
- Here, \(a^2 = 5\) and \(b^2 = 1\).
- Therefore, the equation of the director circle is:

\[ x^2 + y^2 = 5 - 1 = 4 \]

This simplifies to:

\[ x^2 + y^2 = 4 \]

### Step 2: Parametric equations for the ellipse
The ellipse \(\frac{x^2}{2} + \frac{y^2}{10} = 1\) can be represented parametrically as:

\[ x = \sqrt{2} \cos \theta \]
\[ y = \sqrt{10} \sin \theta \]

### Step 3: Substitute parametric equations into the director circle equation
Substituting the parametric equations into the director circle equation \(x^2 + y^2 = 4\):

\[ (\sqrt{2} \cos \theta)^2 + (\sqrt{10} \sin \theta)^2 = 4 \]

This simplifies to:

\[ 2 \cos^2 \theta + 10 \sin^2 \theta = 4 \]

### Step 4: Rearranging the equation
Rearranging the equation gives:

\[ 2 \cos^2 \theta + 10 (1 - \cos^2 \theta) = 4 \]

Expanding this results in:

\[ 2 \cos^2 \theta + 10 - 10 \cos^2 \theta = 4 \]

Combining like terms:

\[ -8 \cos^2 \theta + 10 = 4 \]

### Step 5: Solving for \(\cos^2 \theta\)
Subtracting 10 from both sides:

\[ -8 \cos^2 \theta = -6 \]

Dividing by -8:

\[ \cos^2 \theta = \frac{6}{8} = \frac{3}{4} \]

Taking the square root gives:

\[ \cos \theta = \pm \frac{\sqrt{3}}{2} \]

### Step 6: Finding \(\sin^2 \theta\)
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):

\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4} \]

Taking the square root gives:

\[ \sin \theta = \pm \frac{1}{2} \]

### Step 7: Finding the coordinates \(x\) and \(y\)
Now substituting the values of \(\cos \theta\) and \(\sin \theta\) back into the parametric equations for \(x\) and \(y\):

1. For \(x\):

\[ x = \sqrt{2} \cos \theta = \sqrt{2} \left(\pm \frac{\sqrt{3}}{2}\right) = \pm \frac{\sqrt{6}}{2} \]

2. For \(y\):

\[ y = \sqrt{10} \sin \theta = \sqrt{10} \left(\pm \frac{1}{2}\right) = \pm \frac{\sqrt{10}}{2} \]

### Final Points
Thus, the points from which perpendicular tangents can be drawn to the hyperbola are:

\[ \left(\frac{\sqrt{6}}{2}, \frac{\sqrt{10}}{2}\right), \left(\frac{\sqrt{6}}{2}, -\frac{\sqrt{10}}{2}\right), \left(-\frac{\sqrt{6}}{2}, \frac{\sqrt{10}}{2}\right), \left(-\frac{\sqrt{6}}{2}, -\frac{\sqrt{10}}{2}\right) \]