If `y = mx + c` be a tangent to the hyperbola `x^2/lambda^2-y^2/(lambda^3+lambda^2+lambda)^2 = 1, (lambda!=0),` then minimum value of 16m^2

Options

A. 0
B. 1
C. 4
D. 9

Answer

To solve the problem, we need to find the minimum value of \(16m^2\) given that the line \(y = mx + c\) is a tangent to the hyperbola

\[ \frac{x^2}{\lambda^2} - \frac{y^2}{(\lambda^3 + \lambda^2 + \lambda)^2} = 1 \]

where \(\lambda \neq 0\).

### Step-by-Step Solution:

1. **Identify Parameters of the Hyperbola:**
The hyperbola can be represented in the standard form where \(a^2 = \lambda^2\) and \(b^2 = (\lambda^3 + \lambda^2 + \lambda)^2\).

2. **Condition for Tangency:**
For the line \(y = mx + c\) to be a tangent to the hyperbola, the condition of tangency must hold:
\[ c^2 = a^2 m^2 - b^2 \]
Substituting the values of \(a^2\) and \(b^2\):
\[ c^2 = \lambda^2 m^2 - (\lambda^3 + \lambda^2 + \lambda)^2 \]

3. **Rearranging the Condition:**
Rearranging gives:
\[ c^2 = \lambda^2 m^2 - (\lambda^3 + \lambda^2 + \lambda)^2 \]
Since \(c^2\) must be non-negative, we have:
\[ \lambda^2 m^2 - (\lambda^3 + \lambda^2 + \lambda)^2 \geq 0 \]
This implies:
\[ \lambda^2 m^2 \geq (\lambda^3 + \lambda^2 + \lambda)^2 \]

4. **Solving for \(m^2\):**
Dividing both sides by \(\lambda^2\) (since \(\lambda \neq 0\)):
\[ m^2 \geq \frac{(\lambda^3 + \lambda^2 + \lambda)^2}{\lambda^2} \]
Simplifying the right-hand side:
\[ m^2 \geq \left(\lambda + 1 + \frac{1}{\lambda}\right)^2 \]

5. **Finding the Minimum Value:**
Let \(f(\lambda) = \lambda + 1 + \frac{1}{\lambda}\). To minimize \(f(\lambda)\), we can use calculus or AM-GM inequality. The minimum occurs when \(\lambda = 1\):
\[ f(1) = 1 + 1 + 1 = 3 \]
Thus,
\[ m^2 \geq 3^2 = 9 \]

6. **Calculating \(16m^2\):**
Therefore,
\[ 16m^2 \geq 16 \times 9 = 144 \]

7. **Final Result:**
The minimum value of \(16m^2\) is \(144\).

### Conclusion:
The minimum value of \(16m^2\) is \(144\).