Given. $2 a_{1}=2 \sin \theta$
$\Rightarrow \quad a_{1}=\sin \theta$
and $3 x^{2}+4 y^{2}=12$
$\Rightarrow \quad \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$
Here, $a^{2}=4$ and $b^{2}=3$
$\therefore$
$b^{2}=a^{2}\left(1-e^{2}\right)$
$\Rightarrow$
$3=4\left(1-e^{2}\right)$
$\Rightarrow$
$e^{2}=1-\frac{3}{4}=\frac{1}{4}$
$e=\frac{1}{2}$
Focus, $F(a e, 0)=F\left(2 \times \frac{1}{2}, 0\right)$
$=F(1,0)$
For hyperbola foci are same
$a_{1} e_{1}=a e=1$
$\therefore$
$(\sin \theta) e_{1}=1$
$\Rightarrow$
$e_{1}=\operatorname{cosec} \theta$
and $b_{1}^{2}=a_{1}^{2}\left(e_{1}^{2}-1\right)=a_{1}^{2} e_{1}^{2}-a_{1}^{2}$
$\Rightarrow \quad b_{1}^{2}=1-\sin ^{2} \theta=\cos ^{2} \theta$
$\frac{x^{2}}{a_{1}^{2}}-\frac{y^{2}}{b_{1}^{2}}=1$
$\Rightarrow \quad \frac{x^{2}}{\sin ^{2} \theta}-\frac{y^{2}}{\cos ^{2} \theta}=1$
$\Rightarrow x^{2} \operatorname{cosec}^{2} \theta-y^{2} \sec ^{2} \theta=1$