Let $\mathrm{X}$ - axis be the transverse axis and $\mathrm{Y}$ - axis be the conjugate axis of a hyperbola…
Let $\mathrm{X}$ - axis be the transverse axis and $\mathrm{Y}$ - axis be the conjugate axis of a hyperbola $\mathrm{H}$. Let the eccentricity of $\mathrm{H}$ be the reciprocal of the eccentricity of the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$. If $(5,4)$ is a point on $H$, then the length of the transverse axis of $\mathrm{H}$ is
$2 \sqrt{2}$
4
6
10
Solution
Eccentricity of ellipse $\frac{x^2}{4}+\frac{y^2}{2}$ is:
$e^{\prime}=\sqrt{1-\frac{2}{4}}=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$
Now eccentricity of hyperbola:
$\begin{aligned} & \mathrm{e}=\frac{1}{e^{\prime}}=\sqrt{2} \\ & \Rightarrow \sqrt{1+\frac{b^2}{a^2}}=\sqrt{2} \Rightarrow 1+\frac{b^2}{a^2}=2 \\ & \Rightarrow \frac{b^2}{a^2}=1 \Rightarrow b^2=a^2\end{aligned}$
The equation of hyperbola is:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{a^2}=1$ ...(i)
Since $(5,4)$ lies on equation (i)
So, $\frac{25}{a^2}-\frac{16}{a^2}=1 \Rightarrow 9=a^2 \quad \Rightarrow a= \pm 3$
Now length of transverse axis $=2|a|=2 \times 3=6$