The point `(alpha^(2)+2lamda+5,lamda^(2)+1)` lies on the line `x+y=10` for:

Options

A. All real value of `lamda`
B. Some real value of `lamda`
C. `lamda=-1`
D. `lamda=2`

Answer

To determine the values of \( \lambda \) for which the point \( ( \alpha^2 + 2\lambda + 5, \lambda^2 + 1 ) \) lies on the line \( x + y = 10 \), we can follow these steps:

### Step 1: Substitute the point into the line equation
The point \( (x, y) \) is given as \( ( \alpha^2 + 2\lambda + 5, \lambda^2 + 1 ) \). According to the line equation \( x + y = 10 \), we can substitute:

\[ \alpha^2 + 2\lambda + 5 + \lambda^2 + 1 = 10 \]

### Step 2: Simplify the equation
Combine like terms:

\[ \alpha^2 + 2\lambda + \lambda^2 + 6 = 10 \]

Subtract 10 from both sides:

\[ \alpha^2 + 2\lambda + \lambda^2 - 4 = 0 \]

### Step 3: Rearrange the equation
Rearranging gives us:

\[ \lambda^2 + 2\lambda + \alpha^2 - 4 = 0 \]

### Step 4: Identify the quadratic equation
This is a quadratic equation in \( \lambda \):

\[ \lambda^2 + 2\lambda + (\alpha^2 - 4) = 0 \]

### Step 5: Calculate the discriminant
For the quadratic equation \( ax^2 + bx + c = 0 \), the discriminant \( D \) is given by:

\[ D = b^2 - 4ac \]

Here, \( a = 1 \), \( b = 2 \), and \( c = \alpha^2 - 4 \). Thus,

\[ D = 2^2 - 4 \cdot 1 \cdot (\alpha^2 - 4) \]
\[ D = 4 - 4(\alpha^2 - 4) \]
\[ D = 4 - 4\alpha^2 + 16 \]
\[ D = 20 - 4\alpha^2 \]

### Step 6: Determine the conditions for real values of \( \lambda \)
For \( \lambda \) to have real solutions, the discriminant must be non-negative:

\[ 20 - 4\alpha^2 \geq 0 \]

### Step 7: Solve for \( \alpha^2 \)
Rearranging gives:

\[ 20 \geq 4\alpha^2 \]
\[ 5 \geq \alpha^2 \]

Thus, we have:

\[ \alpha^2 \leq 5 \]

### Conclusion
The point \( ( \alpha^2 + 2\lambda + 5, \lambda^2 + 1 ) \) lies on the line \( x + y = 10 \) for values of \( \lambda \) when \( \alpha^2 \) is less than or equal to 5.

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