If the line x-1=0, is a directrix of the hype | vedclass

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Question

$frac{x^{2}}{6 / k }-\frac{y^{2}}{6}=1$

$e ^{2}=1+\frac{6}{6 / k }$

$e =\sqrt{1+ k }$

$a =\sqrt{\frac{6}{ k }}$

Eq. of directrix $x=\frac{

Vedclass · Accepted Answer

$(\sqrt{5},-2)$

Vedclass · Answer

$frac{x^{2}}{6 / k }-\frac{y^{2}}{6}=1$

$e ^{2}=1+\frac{6}{6 / k }$

$e =\sqrt{1+ k }$

$a =\sqrt{\frac{6}{ k }}$

Eq. of directrix $x=\frac{a}{e} \Rightarrow x=\sqrt{\frac{6}{k(k+1)}}$

$\frac{6}{k(k+1)}=1$

$k=2$

From eq.$(1)$, we get $2 x^{2}-y^{2}=6$

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If the line $x-1=0$, is a directrix of the hyperbola $kx ^{2}- y ^{2}=6$, then the hyperbola passes through the point.

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