If $(1,2)$ is the focus, $x+2 y=0$ is the directrix and $\sqrt{2}$ is the eccentricity of a hyperbola, then…
If $(1,2)$ is the focus, $x+2 y=0$ is the directrix and $\sqrt{2}$ is the eccentricity of a hyperbola, then the equation of the hyperbola is
- $x^2-y^2=a^2$
- $3 x^2-8 x y-3 y^2-10 x-20 y+25=0$
- $x y=c^2$
- $3 x^2-8 x y-3 y^2+10 x-20 y-25=0$
Solution
Focus is $(1,2)$, directrix is $x+2 y=0, e=\sqrt{2}$
$\therefore$ Hyperbola is
$
\begin{aligned}
& (x-1)^2+(y-2)^2=(\sqrt{2})^2 \frac{(x+2 y)^2}{1+4} \\
& \Rightarrow x^2+1-2 x+y^2+4-4 y=\frac{2}{5}\left[x^2+4 y^2+4 x y\right] \\
& \Rightarrow 5 x^2-10 x+5 y^2-20 y+25=2 x^2+8 y^2+8 x y \\
& \Rightarrow 3 x^2-8 x y-3 y^2-10 x-20 y+25=0
\end{aligned}
$
Asked in: AP EAMCET 2022 (07 Jul Shift 2)
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