Q.107. Parabola.

Question 107.
The parabola y^2 = kx makes an intercept of length 4 cm on the line x - 2y = 1. Find the value of k.

Answer 107.
If the points of intersection are (x1, y1) and (x2, y2),
putting in the equation of the line, x - 2y = 1
=> (x1 - x2) - 2(y1 - y2) = 0
=> (x1 - x2) = 2(y1 - y2)
=> length of intercept
= √[(x1 - x2)^2 + (y1 - y2)^2]
= √[4(y1 - y2)^2 + (y1 - y2)^2]
= √5 * ly1 - y2l ... (1)

Solving y^2 = kx with x - 2y = 1,
y^2 - k(2y + 1) = 0
=> y^2 - 2ky - k = 0
=> y = (1/2) [2k ± √(4k^2 + 4k)]
=> y = k ± √(k^2 + k)
=> l y1 - y2 l = 2√(k^2 + k) ... (2)

From (1) and (2),
length of intercept,
4 = 2√[5(k^2 + k)]
=> 16 = 20(k^2 + k)
=> 5k^2 + 5k - 4 = 0
=> k = (1/10) [- 5 ± √(105)]

k cannot be negative because in that case the parabola will be in the third and the fourth quadrant and the line x - 2y = 1 is in 1st, 3rd and fourth quadrant which will nlot intersect the parabola with negative k in two points.

=> k is positive and its value is
(√(105) - 5)/10.

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