The value of square of slope of the common tangent to the curves `4x^2+9y^2=36 and (2x)^2 +(2y)^2=31`

Answer

To find the value of the square of the slope of the common tangent to the curves \(4x^2 + 9y^2 = 36\) and \((2x)^2 + (2y)^2 = 31\), we can follow these steps:

### Step 1: Rewrite the equations of the curves
First, we rewrite the equations of the curves in standard form.

1. The first curve is an ellipse:
\[ 4x^2 + 9y^2 = 36 \]
Dividing through by 36 gives:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
Here, \(a^2 = 9\) and \(b^2 = 4\).

2. The second curve is a circle:
\[ (2x)^2 + (2y)^2 = 31 \]
Simplifying gives:
\[ 4x^2 + 4y^2 = 31 \quad \Rightarrow \quad x^2 + y^2 = \frac{31}{4} \]

### Step 2: Find the equation of the tangent to the ellipse
The equation of the tangent to the ellipse in slope form \(y = mx + c\) can be expressed as:
\[ y = mx \pm \sqrt{a^2m^2 + b^2} \]
Substituting \(a^2 = 9\) and \(b^2 = 4\):
\[ y = mx \pm \sqrt{9m^2 + 4} \]

### Step 3: Determine the radius of the circle
For the circle \(x^2 + y^2 = \frac{31}{4}\), the radius \(r\) is:
\[ r = \sqrt{\frac{31}{4}} = \frac{\sqrt{31}}{2} \]

### Step 4: Use the condition for the common tangent
The distance from the center of the circle (which is at the origin \((0,0)\)) to the tangent line must equal the radius. The distance \(d\) from the point \((0,0)\) to the line \(y = mx + c\) is given by:
\[ d = \frac{|c|}{\sqrt{1 + m^2}} \]
Setting this equal to the radius:
\[ \frac{\sqrt{9m^2 + 4}}{\sqrt{1 + m^2}} = \frac{\sqrt{31}}{2} \]

### Step 5: Square both sides and simplify
Squaring both sides gives:
\[ \frac{9m^2 + 4}{1 + m^2} = \frac{31}{4} \]
Cross-multiplying results in:
\[ 4(9m^2 + 4) = 31(1 + m^2) \]
Expanding both sides:
\[ 36m^2 + 16 = 31 + 31m^2 \]
Rearranging yields:
\[ 5m^2 = 15 \quad \Rightarrow \quad m^2 = 3 \]

### Step 6: Conclusion
Thus, the value of the square of the slope of the common tangent is:
\[ \boxed{3} \]