Question:medium

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:

Show 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 \(\text{L}_1 < \text{L}_2\)!
Updated On: May 29, 2026
  • \( \text{L}_1 > \text{L}_2 \)
  • \( \text{L}_1 = \text{L}_2 \)
  • \( \text{L}_1 < \text{L}_2 \)
  • \( \text{L}_1/\text{L}_2 = \sqrt{2} \)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
This question is from coordinate geometry and involves finding the intersection points of a circle and a parabola to determine the length of their common chord.
We need to calculate the length of the common chord ($L_1$), find the latus rectum of the parabola ($L_2$), and then compare their values.
Step 2: Key Formulas and Approach:
Latus Rectum of a Parabola: For a parabola $y^2 = 4ax$, the length of the latus rectum is $4a$.

Common Chord Length: Solve the equations of the circle $x^2 + y^2 = 9$ and the parabola $y^2 = 8x$ simultaneously to find their points of intersection. The distance between these points along the vertical line is the length of the common chord.

Step 3: Detailed Explanation:

Calculate $L_2$:
The equation of the parabola is $y^2 = 8x$.
Comparing this with $y^2 = 4ax$, the length of the latus rectum is:
\[ L_2 = 8 \quad \cdots (1) \]
Find the Intersection Points:
Substitute the parabola equation $y^2 = 8x$ into the circle equation $x^2 + y^2 = 9$:
\[ x^2 + 8x = 9 \quad \implies \quad x^2 + 8x - 9 = 0 \] Factoring the quadratic equation:
\[ (x + 9)(x - 1) = 0 \quad \implies \quad x = 1 \text{ or } x = -9 \] Since $y^2 = 8x$, a negative $x$-value ($x = -9$) would yield imaginary values for $y$. Thus, we choose the valid real solution $x = 1$.

Calculate the Coordinates:
For $x = 1$:
\[ y^2 = 8(1) = 8 \quad \implies \quad y = \pm\sqrt{8} = \pm2\sqrt{2} \] The intersection points are $(1, 2\sqrt{2})$ and $(1, -2\sqrt{2})$.

Calculate the Chord Length ($L_1$):
The length of the common chord is the distance between these two points:
\[ L_1 = 2\sqrt{2} - (-2\sqrt{2}) = 4\sqrt{2}\text{ units} \] Approximating the value:
\[ L_1 = 4 \times 1.414 = 5.66\text{ units} \]
Compare $L_1$ and $L_2$:
Comparing our results, we have $L_1 = 5.66$ and $L_2 = 8$.
Since $5.66<8$, we conclude that $L_1<L_2$.

Step 4: Final Answer:
The correct relationship is $L_1<L_2$, which corresponds to Option (C).
Was this answer helpful?
0