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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $2 x^{2}-y^{2}=6$

Check options

If the line $x-1=0$, is a directrix of the hyperbola $kx ^{2}- y ^{2}=6$, then the hyperbola passes through the point.

Similar Questions

The distance between the directrices of the hyperbola $x = 8\sec \theta ,\;\;y = 8\tan \theta $ is

The point of contact of the tangent $y = x + 2$ to the hyperbola $5{x^2} - 9{y^2} = 45$ is

What will be equation of that chord of hyperbola $25{x^2} - 16{y^2} = 400$, whose mid point is $(5, 3)$

The chord $ PQ $ of the rectangular hyperbola $xy = a^2$ meets the axis of $x$ at $A ; C $ is the mid point of $ PQ\ \& 'O' $ is the origin. Then the $ \Delta ACO$ is :

The equation of the hyperbola in the standard form (with transverse axis along the $x$ - axis) having the length of the latus rectum = $9$ units and eccentricity = $5/4$ is