If $y=x+\sqrt{2}$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{2}=1$, then equations of its…
If $y=x+\sqrt{2}$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{2}=1$, then equations of its directrices are
$x= \pm \sqrt{3}$
$x= \pm \sqrt{\frac{8}{3}}$
$x= \pm \sqrt{\frac{2}{3}}$
$x= \pm \sqrt{\frac{4}{3}}$
Solution
Hyperbola: $\frac{x^2}{a^2}-\frac{y^2}{2}=1$. Now, equation of tangent
in slope form $y=m x \pm \sqrt{a^2 m^2-2}$
Comparing with given tangent line $y=x+\sqrt{2}$ We get, $m=1$ and $a^2-2=2 \Rightarrow a^2=4$
$e=\sqrt{1+\frac{2}{4}}=\sqrt{\frac{3}{2}} .$ So, equation of directrix $x= \pm \frac{a}{e}= \pm \sqrt{\frac{8}{3}}$.