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

Question

If the line x-1=0, is a directrix of the hyperbola kx 2- y 2=6, then the hyperbola passes through the point (A) (-2 √5, 6) (B) (-√5, 3) (C) (√5,-2) (D) (2 √5, 3 √6)

Tardigrade Admin · Accepted Answer

(x2/6 / k)-(y2/6)=1 .....(1) e2=1+(6/6 / k) e=√1+k a=√(6/k) Eq. of directrix x=(a/e) ⇒ x=√(6/k(k+1)) (6/ k ( k +1))=1 k =2 From eq. (1), we get 2 x2-y2=6 Check options

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

$(- 25, 6)$

$(- 5, 3)$

$(5, - 2)$

$(25, 36)$

$\frac{x ^{2}}{6/ k} - \frac{y ^{2}}{6} = 1.....$ (1)
$e^{2} = 1 + \frac{6}{6/ k}$
$e = 1 + k$
$a = \frac{6}{k}$
Eq. of directrix $x = \frac{a}{e} \Rightarrow x = \frac{6}{k ( k + 1 )}$
$\frac{6}{k ( k + 1 )} = 1$
$k = 2$
From eq. (1), we get $2 x^{2} - y^{2} = 6$
Check options

Q. If the line x−1=0, is a directrix of the hyperbola kx2−y2=6, then the hyperbola passes through the point

(−25​,6)

(−5​,3)

(5​,−2)

(25​,36​)