The points of contact of the tangents drawn from the point (4, 6) to the parabola `y^(2) = 8x`

Options

A. (2, 4), (18, 12)
B. (2, 4), (8, 8)
C. (8,8) (18,12)
D. (0,0) `(1,2sqrt(2))`

Answer

To find the points of contact of the tangents drawn from the point (4, 6) to the parabola \( y^2 = 8x \), we can follow these steps:

### Step 1: Identify the standard form of the parabola
The given parabola is \( y^2 = 8x \). We can compare this with the standard form \( y^2 = 4ax \) to find \( a \):
\[ 4a = 8 \implies a = 2 \]

### Step 2: Write the equation of the tangent
The equation of the tangent to the parabola \( y^2 = 8x \) at the point of contact can be expressed as:
\[ y = mx + \frac{2}{m} \]
where \( m \) is the slope of the tangent.

### Step 3: Substitute the point (4, 6) into the tangent equation
Since the point (4, 6) lies on the tangent, we can substitute \( x = 4 \) and \( y = 6 \) into the tangent equation:
\[ 6 = 4m + \frac{2}{m} \]

### Step 4: Multiply through by \( m \) to eliminate the fraction
Multiplying both sides by \( m \) gives:
\[ 6m = 4m^2 + 2 \]

### Step 5: Rearrange the equation
Rearranging the equation leads to:
\[ 4m^2 - 6m + 2 = 0 \]

### Step 6: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 4 \), \( b = -6 \), and \( c = 2 \).
\[ m = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot 2}}{2 \cdot 4} \]
\[ m = \frac{6 \pm \sqrt{36 - 32}}{8} \]
\[ m = \frac{6 \pm \sqrt{4}}{8} \]
\[ m = \frac{6 \pm 2}{8} \]
This gives us two values:
\[ m_1 = \frac{8}{8} = 1 \quad \text{and} \quad m_2 = \frac{4}{8} = \frac{1}{2} \]

### Step 7: Find the points of contact for each slope
1. **For \( m = 1 \)**:
\[ y = x + 2 \]
Substitute into the parabola \( y^2 = 8x \):
\[ (x + 2)^2 = 8x \]
\[ x^2 + 4x + 4 = 8x \]
\[ x^2 - 4x + 4 = 0 \]
\[ (x - 2)^2 = 0 \implies x = 2 \]
Then, \( y = 2 + 2 = 4 \). So, one point of contact is \( (2, 4) \).

2. **For \( m = \frac{1}{2} \)**:
\[ y = \frac{1}{2}x + 4 \]
Substitute into the parabola:
\[ \left(\frac{1}{2}x + 4\right)^2 = 8x \]
\[ \frac{1}{4}x^2 + 4x + 16 = 8x \]
\[ \frac{1}{4}x^2 - 4x + 16 = 0 \]
Multiply through by 4:
\[ x^2 - 16x + 64 = 0 \]
\[ (x - 8)^2 = 0 \implies x = 8 \]
Then, \( y = \frac{1}{2}(8) + 4 = 8 \). So, the other point of contact is \( (8, 8) \).

### Final Answer
The points of contact of the tangents drawn from the point (4, 6) to the parabola \( y^2 = 8x \) are \( (2, 4) \) and \( (8, 8) \).

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