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Question:

Let L1 be the length of the common chord of the curves x² + y² = 9 and y² = 8x, and L2 be the length of the latus rectum of y² = 8x, then :

L1>L2

L1<L2

L1L2=√2

L1=L2

Solution:

The given curves are x² + y² = 9 and y² = 8x. To find the points of intersection, let us substitute the values of y².
x² + 8x − 9 = 0
(x + 9)(x − 1) = 0
x = −9 is not possible.
Hence, x = 1
y² = 8 ⇒ y = ±2√2
The two points are (1, 2√2) and (1, −2√2)
L1 = 2 × 2√2 = 4√2
L2 = 4a = 8 (where a is the distance from the vertex to the focus in the parabola y² = 4ax)
Hence, L2 > L1