The exhaustive set of values of k for which tangents drawn from the point (k + 3, k) to the parabola `y^(2)=4x`, are real, is

Options

A. (-2, 6)
B. `(-oo,-2)cup(6,oo)`
C. `(-oo,-2]cup[6,oo)`
D. None of these

Answer

To solve the problem of finding the exhaustive set of values of \( k \) for which the tangents drawn from the point \( (k + 3, k) \) to the parabola \( y^2 = 4x \) are real, we can follow these steps:

### Step 1: Understand the condition for tangents to be real
For the tangents from a point to a parabola to be real, the point must lie outside the parabola. The condition for this is given by the equation \( S_1 > 0 \), where \( S_1 \) is derived from the equation of the parabola.

### Step 2: Write the equation of the parabola
The equation of the parabola is \( y^2 = 4x \). We can rewrite this in the form \( S = y^2 - 4x = 0 \).

### Step 3: Substitute the point into the parabola's equation
We substitute the point \( (k + 3, k) \) into the equation:
\[ S_1 = k^2 - 4(k + 3) \]
This simplifies to:
\[ S_1 = k^2 - 4k - 12 \]

### Step 4: Set up the inequality
For the tangents to be real, we need:
\[ k^2 - 4k - 12 > 0 \]

### Step 5: Factor the quadratic inequality
We can factor the quadratic expression:
\[ k^2 - 4k - 12 = (k - 6)(k + 2) \]
Thus, we need to solve the inequality:
\[ (k - 6)(k + 2) > 0 \]

### Step 6: Determine the intervals
To find the intervals where this product is positive, we can find the critical points by setting each factor to zero:
- \( k - 6 = 0 \) gives \( k = 6 \)
- \( k + 2 = 0 \) gives \( k = -2 \)

Now we test the intervals:
1. \( k < -2 \)
2. \( -2 < k < 6 \)
3. \( k > 6 \)

### Step 7: Test the intervals
- For \( k < -2 \) (e.g., \( k = -3 \)):
\((k - 6)(k + 2) = (-3 - 6)(-3 + 2) = (-9)(-1) > 0\) (True)

- For \( -2 < k < 6 \) (e.g., \( k = 0 \)):
\((0 - 6)(0 + 2) = (-6)(2) < 0\) (False)

- For \( k > 6 \) (e.g., \( k = 7 \)):
\((7 - 6)(7 + 2) = (1)(9) > 0\) (True)

### Step 8: Write the solution
Thus, the values of \( k \) for which the tangents are real are:
\[ k \in (-\infty, -2) \cup (6, \infty) \]

### Final Answer
The exhaustive set of values of \( k \) is \( (-\infty, -2) \cup (6, \infty) \).

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