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  • If the line  is the directrix of the parabola  <img alt="y^{2}-kx+8=0" src="https://entranc

If the line x-1=0 is the directrix of the parabola  y^{2}-kx+8=0  , then one of the values of k is

Option: 1

\frac{1}{8}


Option: 2

8


Option: 3

4


Option: 4

\frac{1}{4}


Answers (1)

best_answer

Given,

y^{2}-kx+8=0

\Rightarrow y^2=k\left(x-\frac{8}{k}\right)

Shifting the origin Y^{2}=kX where   Y=y, X=\left(x-\frac{8}{k}\right)

Directrix of standard parabola is  X=-\frac{k}{4}

 Directrix of original parabola is  X=\frac{8}{k}-\frac{k}{4}

Now, x = 1 also coincides with  X=\frac{8}{k}-\frac{k}{4}

On solving, we get   k=4

Posted by

Sanket Gandhi

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