The length of the sides of the square which can be made by four perpendicular tangents to the ellipse `x^2/7+(2y^2)/11=1`, is

Answer

To find the length of the sides of the square formed by four perpendicular tangents to the ellipse given by the equation \( \frac{x^2}{7} + \frac{2y^2}{11} = 1 \), we can follow these steps:

### Step 1: Identify the semi-major and semi-minor axes
The equation of the ellipse can be rewritten in the standard form:
\[ \frac{x^2}{7} + \frac{y^2}{\frac{11}{2}} = 1 \]
From this, we can identify:
- \( a^2 = 7 \) (where \( a \) is the semi-major axis)
- \( b^2 = \frac{11}{2} \) (where \( b \) is the semi-minor axis)

### Step 2: Determine the slopes of the tangents
For the square formed by the tangents, the slopes of the tangents will be \( m = 1 \) and \( m = -1 \) (since they are perpendicular).

### Step 3: Write the equations of the tangents
The general equation of the tangent to the ellipse is given by:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]
Substituting \( m = 1 \):
\[ y = x \pm \sqrt{7 \cdot 1^2 + \frac{11}{2}} = x \pm \sqrt{7 + \frac{11}{2}} = x \pm \sqrt{\frac{14 + 11}{2}} = x \pm \sqrt{\frac{25}{2}} = x \pm \frac{5}{\sqrt{2}} \]
This gives us two tangent lines:
1. \( y = x + \frac{5}{\sqrt{2}} \)
2. \( y = x - \frac{5}{\sqrt{2}} \)

Now substituting \( m = -1 \):
\[ y = -x \pm \sqrt{7 \cdot (-1)^2 + \frac{11}{2}} = -x \pm \sqrt{7 + \frac{11}{2}} = -x \pm \frac{5}{\sqrt{2}} \]
This gives us two more tangent lines:
1. \( y = -x + \frac{5}{\sqrt{2}} \)
2. \( y = -x - \frac{5}{\sqrt{2}} \)

### Step 4: Find the points of intersection of the tangents
To find the vertices of the square, we need to find the intersection points of these lines.

1. Intersection of \( y = x + \frac{5}{\sqrt{2}} \) and \( y = -x + \frac{5}{\sqrt{2}} \):
\[ x + \frac{5}{\sqrt{2}} = -x + \frac{5}{\sqrt{2}} \implies 2x = 0 \implies x = 0 \implies y = \frac{5}{\sqrt{2}} \]
So one vertex is \( (0, \frac{5}{\sqrt{2}}) \).

2. Intersection of \( y = x - \frac{5}{\sqrt{2}} \) and \( y = -x - \frac{5}{\sqrt{2}} \):
\[ x - \frac{5}{\sqrt{2}} = -x - \frac{5}{\sqrt{2}} \implies 2x = -\frac{10}{\sqrt{2}} \implies x = -\frac{5}{\sqrt{2}} \implies y = -\frac{5}{\sqrt{2}} \]
So another vertex is \( (-\frac{5}{\sqrt{2}}, 0) \).

### Step 5: Calculate the length of the side of the square
The distance between the points \( (0, \frac{5}{\sqrt{2}}) \) and \( (-\frac{5}{\sqrt{2}}, 0) \) can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{\left(-\frac{5}{\sqrt{2}} - 0\right)^2 + \left(0 - \frac{5}{\sqrt{2}}\right)^2} = \sqrt{\left(-\frac{5}{\sqrt{2}}\right)^2 + \left(-\frac{5}{\sqrt{2}}\right)^2} \]
\[ = \sqrt{\frac{25}{2} + \frac{25}{2}} = \sqrt{25} = 5 \]

### Conclusion
The length of the sides of the square formed by the four perpendicular tangents to the ellipse is \( 5 \) units.