Let `L_(1)` be the length of the common chord of the curves `x^(2)+y^(2)=9 and y^(2)=8x and L_(2)` be the length of the latus rectum of `y^(2)=8x`, then

Options

A. `L_(1) gt L_(2)`
B. `L_(1)=L_(2)`
C. `L_(1) lt L_(2)`
D. `(L_(1))/(L_(2))=sqrt2`

Answer

To solve the problem, we need to find the lengths \( L_1 \) and \( L_2 \) as defined in the question.

### Step 1: Determine \( L_2 \) - Length of the Latus Rectum of the Parabola \( y^2 = 8x \)

The equation \( y^2 = 8x \) represents a parabola that opens to the right. The standard form of a parabola is \( y^2 = 4ax \), where \( a \) is the distance from the vertex to the focus.

From the equation \( y^2 = 8x \), we can identify \( 4a = 8 \), which gives us \( a = 2 \).

The length of the latus rectum \( L_2 \) of a parabola is given by the formula \( L_2 = 4a \).

Substituting the value of \( a \):
\[ L_2 = 4 \times 2 = 8 \]

### Step 2: Determine \( L_1 \) - Length of the Common Chord of the Circle and the Parabola

The circle is given by the equation \( x^2 + y^2 = 9 \).

To find the points of intersection between the circle and the parabola, we substitute \( y^2 = 8x \) into the circle's equation:
\[ x^2 + 8x = 9 \]
Rearranging gives:
\[ x^2 + 8x - 9 = 0 \]

### Step 3: Solve the Quadratic Equation

To solve for \( x \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 8 \), and \( c = -9 \):
\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} \]
\[ x = \frac{-8 \pm \sqrt{64 + 36}}{2} \]
\[ x = \frac{-8 \pm \sqrt{100}}{2} \]
\[ x = \frac{-8 \pm 10}{2} \]

Calculating the two possible values for \( x \):
1. \( x = \frac{2}{2} = 1 \)
2. \( x = \frac{-18}{2} = -9 \)

### Step 4: Find Corresponding \( y \) Values

For \( x = 1 \):
\[ y^2 = 8 \cdot 1 = 8 \implies y = \pm 2\sqrt{2} \]

For \( x = -9 \):
\[ y^2 = 8 \cdot (-9) \text{ (not valid, since } y^2 \text{ cannot be negative)} \]

Thus, the points of intersection are \( (1, 2\sqrt{2}) \) and \( (1, -2\sqrt{2}) \).

### Step 5: Calculate the Length of the Common Chord \( L_1 \)

The length of the common chord \( L_1 \) can be calculated using the distance formula between the two intersection points:
\[ L_1 = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Here, \( (x_1, y_1) = (1, 2\sqrt{2}) \) and \( (x_2, y_2) = (1, -2\sqrt{2}) \):
\[ L_1 = \sqrt{(1 - 1)^2 + (2\sqrt{2} - (-2\sqrt{2}))^2} \]
\[ L_1 = \sqrt{0 + (4\sqrt{2})^2} = \sqrt{16 \cdot 2} = \sqrt{32} = 4\sqrt{2} \]

### Step 6: Compare \( L_1 \) and \( L_2 \)

Now we have:
- \( L_1 = 4\sqrt{2} \)
- \( L_2 = 8 \)

To compare \( L_1 \) and \( L_2 \):
\[ 4\sqrt{2} \approx 4 \times 1.414 = 5.656 \]
Since \( 5.656 < 8 \), we conclude that \( L_2 > L_1 \).

### Final Answer

Thus, we find that \( L_2 > L_1 \).

---