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:
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: Feb 3, 2025
- \( L_1>L_2 \)
- \( L_1 = L_2 \)
- \( L_1<L_2 \)
- \( \frac{L_1}{L_2} = \sqrt{2} \)
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The Correct Option is C
Solution and Explanation
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:
\[
L_1<L_2
\]
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