Question:medium

Let \(L_1\) be the length of the common chord of the curves \[ x^2 + y^2 = 9 \quad {and} \quad y^2 = 8x \] and let \(L_2\) be the length of the latus rectum of \(y^2 = 8x\). Then:

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The length of the latus rectum of a parabola is always \(4a\), which helps in quick calculations.
Updated On: Jan 13, 2026
  • \( L_1>L_2 \)
  • \( L_1 = L_2 \)
  • \( L_1<L_2 \)
  • \( \frac{L_1}{L_2} = \sqrt{2} \)
Show Solution

The Correct Option is C

Solution and Explanation

The provided equations are:\[x^2 + y^2 = 9\]\[y^2 = 8x\]To find the intersection points, substitute \( y^2 \) from the second equation into the first:\[x^2 + 8x = 9\]Rearranging this equation gives:\[x^2 + 8x - 9 = 0\]Factoring the quadratic equation yields:\[(x + 9)(x - 1) = 0\]This results in \( x = -9 \) or \( x = 1 \).When \( x = 1 \):\[y^2 = 8(1) = 8\]Solving for \( y \):\[y = \pm \sqrt{8} = \pm 2\sqrt{2}\]One intersection point is \( (1, 2\sqrt{2}) \) and another is \( (1, -2\sqrt{2}) \). The common chord connects these two points. The length of the common chord (\( L_1 \)) is the distance between these two points:\[L_1 = \sqrt{(1-1)^2 + (2\sqrt{2} - (-2\sqrt{2}))^2} = \sqrt{0^2 + (4\sqrt{2})^2} = 4\sqrt{2}\]The length of the latus rectum of the parabola \( y^2 = 8x \) is given by \( L_2 = 4a \). For this parabola, \( 4a = 8 \), so \( a = 2 \). Thus, the length of the latus rectum is:\[L_2 = 4a = 4 \times 2 = 8\]Comparing the lengths, \( L_1 = 4\sqrt{2} \approx 5.66 \) and \( L_2 = 8 \):\[L_1<L_2\]
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