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  • If the line   is the directrix of the parabola  <img alt="\mathrm {y^2-k x+8=0

If the line \mathrm {x-1=0}  is the directrix of the parabola  \mathrm {y^2-k x+8=0,} then positive value of \mathrm {k}  is

Option: 1

4


Option: 2

3


Option: 3

1


Option: 4

0


Answers (1)

best_answer

The given parabola is  \mathrm {y^2=k x-8=k\left(x-\frac{8}{k}\right)}
Shifting the origin to  \mathrm {\left(\frac{8}{k}, 0\right)} , equation of the parabola becomes  \mathrm {Y^2=4 X} where \mathrm {X=x-\frac{8}{k} \: and\: y=Y.}
Directrix of this parabola is  \mathrm {X=\frac{-k}{4} \: or\: x-\frac{8}{k}=\frac{-k}{4}}
This will coincident with   \mathrm {x=1, if \frac{8}{k}-\frac{k}{4}=1}
\begin{aligned} & \mathrm {\Rightarrow \quad k^2+4 k-32=0} \\ & \mathrm {\Rightarrow \quad(k+8)(k-4)=0 }\\ &\mathrm { \therefore \quad k=4 \text {, or }-8 \quad \therefore \quad k=4} \end{aligned}

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