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If x = 1 is the directrix of the parabola y2 = kx - 8, then k is:
1. \(\frac{1}{8}\)
2. 8
3. 4
4. \(\frac{1}{4}\)

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Correct Answer - Option 3 : 4

Concept:

The directrix of a parabola whose is equation is of the form (y - q)2 = 4a(x - p), is the line x = p - a.

 

Calculation:

The given equation of the parabola y2 = kx - 8 can be re-written as:

⇒ \(\rm (y-0)^2=4\left(\frac{k}{4}\right)\left(x-\frac{8}{k}\right)\)

Comparing the above equation with the general form of the equation (y - q)2 = 4a(x - p), we have:

q = 0, \(\rm a=\frac k4\), \(\rm p=\frac 8k\).

The equation of the directrix is:

x = p - a

⇒ \(\rm x=\frac 8k-\frac k4\)

According to the question, x = 1 is the directrix.

∴ \(\rm \frac 8k-\frac k4=1\)

⇒ k2 + 4k - 32 = 0

⇒ k2 + 8k - 4k - 32 = 0

⇒ k(k + 8) - 4(k + 8) = 0

⇒ (k + 8)(k - 4) = 0

⇒ k + 8 = 0 OR k - 4 = 0

⇒ k = -8 OR k = 4.

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