Question

Let \(\text{L}_1\) be the length of the common chord of the curves \(x^2 + y^2 = 9\) and \(y^2 = 8x\), and \(\text{L}_2\) be the length of the latus rectum of \(y^2 = 8x\), then:

• A. \( \text{L}_1 > \text{L}_2 \)
• B. \( \text{L}_1 = \text{L}_2 \)
• C. \( \text{L}_1 < \text{L}_2 \)
• D. \( \text{L}_1/\text{L}_2 = \sqrt{2} \)

Hint:

You can visualize this conceptually without doing the arithmetic distance step. The circle has a radius of \(3\), meaning its absolute maximum span anywhere along the vertical axis is its diameter, which is \(6\). Since the latus rectum length is \(8\), the common chord bounded inside this small circle could never possibly be larger than \(6\), instantly proving \(\text{L}_1 < \text{L}_2\)!

Answer 1

The Correct Option is C

Concept: To find the length of a common chord between two intersecting conic curves:
• Find the points of intersection by solving the system of equations simultaneously.
• Identify the corresponding coordinate values at the intersection boundary.
• Compute the distance between these intersection points using the distance formula. For a standard parabola of the form \(y^2 = 4ax\), the length of its latus rectum is directly given by the coefficient constant \(4a\).

Step 1: Finding the length of the latus rectum ($\text{L}_2$).
Given the equation of the parabola: \[ y^2 = 8x \] Comparing this with the standard form \(y^2 = 4ax\), we have \(4a = 8\). Therefore, the length of the latus rectum is: \[ \text{L}_2 = 8 \quad \cdots (1) \]

Step 2: Finding the intersection points of the two curves.
The equations of the two curves are: \[ x^2 + y^2 = 9 \quad \cdots (2) \] \[ y^2 = 8x \quad \cdots (3) \] Substitute the value of \(y^2\) from equation (3) into equation (2): \[ x^2 + 8x = 9 \] Rearranging into a standard quadratic equation format: \[ x^2 + 8x - 9 = 0 \] Splitting the middle term to factor the quadratic equation: \[ x^2 + 9x - x - 9 = 0 \] \[ x(x + 9) - 1(x + 9) = 0 \] \[ (x - 1)(x + 9) = 0 \] This yields two possible values for \(x\): \[ x = 1 \quad \text{or} \quad x = -9 \] Since \(y^2 = 8x\), if \(x = -9\), then \(y^2 = -72\), which gives imaginary values for \(y\). Thus, we discard \(x = -9\). Taking the valid real value \(x = 1\): \[ y^2 = 8(1) = 8 \quad \implies \quad y = \pm \sqrt{8} = \pm 2\sqrt{2} \] Hence, the two intersection points are \(P(1, 2\sqrt{2})\) and \(Q(1, -2\sqrt{2})\).

Step 3: Calculating the length of the common chord ($\text{L}_1$) and comparing values.
The common chord is the straight line segment joining the two intersection points \(P\) and \(Q\). Since both points share the same \(x\)-coordinate (\(x = 1\)), the chord is a vertical line segment, and its length is simply the difference between their \(y\)-coordinates: \[ \text{L}_1 = 2\sqrt{2} - (-2\sqrt{2}) = 4\sqrt{2} \] Approximating the value using \(\sqrt{2} \approx 1.414\): \[ \text{L}_1 = 4 \times 1.414 = 5.656 \text{ units} \] Now, comparing our calculated values: \[ \text{L}_1 = 4\sqrt{2} \approx 5.66 \quad \text{and} \quad \text{L}_2 = 8 \] Since \(5.66 < 8\), we conclude that: \[ \text{L}_1 < \text{L}_2 \]