10-2. Parabola, Ellipse, Hyperbola
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यदि रेखा $x -1=0$, अतिपरवलय $kx ^2- y ^2=6$ की एक नियता है, तो यह अतिपरवलय किस बिंदु से होकर जाता है ?

A

$(-2 \sqrt{5}, 6)$

B

$(-\sqrt{5}, 3)$

C

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

D

$(2 \sqrt{5}, 3 \sqrt{6})$

(JEE MAIN-2022)

Solution

$frac{x^{2}}{6 / k }-\frac{y^{2}}{6}=1$

$e ^{2}=1+\frac{6}{6 / k }$

$e =\sqrt{1+ k }$

$a =\sqrt{\frac{6}{ k }}$

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

Standard 11
Mathematics

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