Question

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:

• A. \( L_1>L_2 \)
• B. \( L_1 = L_2 \)
• C. \( L_1<L_2 \)
• D. \( \frac{L_1}{L_2} = \sqrt{2} \)

Hint:

The length of the latus rectum of a parabola is always \(4a\), which helps in quick calculations.

Answer 1

The Correct Option is C

The given equations are: \[ x^2 + y^2 = 9 \] \[ y^2 = 8x \] Solving for the intersection points: \[ x^2 + 8x = 9 \] Rearrange: \[ x^2 + 8x - 9 = 0 \] Factoring: \[ (x + 9)(x - 1) = 0 \] So, \( x = -9, 1 \). For \( x = 1 \): \[ y^2 = 8(1) = 8 \] \[ y = \pm 2\sqrt{2} \] Thus, length of the common chord: \[ L_1 = \sqrt{(2\sqrt{2})^2 + (2\sqrt{2})^2} = 4\sqrt{2} \] Now, the length of the latus rectum of the parabola \( y^2 = 8x \) is: \[ L_2 = 4a = 4 \times 2 = 8 \] Since \( L_1 = 4\sqrt{2} \approx 5.66 \) and \( L_2 = 8 \), we get: \[ L_1<L_2 \]

Answer 2

Let \(L_1\) be the length of the common chord of the curves
\[ x^2 + y^2 = 9 \quad \text{and} \quad y^2 = 8x \] To find \(L_1\):
1. The equation of the circle is \(x^2 + y^2 = 9\), which represents a circle with center at \((0, 0)\) and radius 3.
2. The equation of the parabola is \(y^2 = 8x\), which represents a parabola with its vertex at \((0, 0)\) and opening along the x-axis.
To find the common chord, we solve the two equations simultaneously. Substituting \(y^2 = 8x\) into \(x^2 + y^2 = 9\), we get: \[ x^2 + 8x = 9 \] \[ x^2 + 8x - 9 = 0 \] Using the quadratic formula to solve for \(x\): \[ x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-9)}}{2(1)} = \frac{-8 \pm \sqrt{64 + 36}}{2} = \frac{-8 \pm \sqrt{100}}{2} = \frac{-8 \pm 10}{2} \] This gives \(x = 1\) or \(x = -9\). These are the x-coordinates of the points of intersection of the circle and the parabola.
Now, substituting these values of \(x\) back into the equation \(y^2 = 8x\) gives the corresponding values of \(y\). For \(x = 1\), \(y^2 = 8\), so \(y = \pm \sqrt{8}\). For \(x = -9\), \(y^2 = -72\), which has no real solutions.
Thus, the common chord is the line joining the points \((1, \sqrt{8})\) and \((1, -\sqrt{8})\), and the length of this chord is: \[ L_1 = 2 \times \sqrt{8} = 4\sqrt{2} \] To find \(L_2\):
The length of the latus rectum of the parabola \(y^2 = 8x\) is given by the formula \(L_2 = \frac{4a}{b^2}\), where \(a = 2\) is the focal length. The length of the latus rectum is: \[ L_2 = \frac{4 \times 2}{8} = 1 \] Thus, we have \(L_1 = 4\sqrt{2}\) and \(L_2 = 1\). Since \(4\sqrt{2} > 1\), the correct relation is:

\( L_1 > L_2 \)