Consider a branch of the hyperbola $x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0$ with vertex at the point $A$. Let $B$ be one of the end points of its latus rectum. If $C$ is the focus of the hyperbola nearest to the point $A$, then the area of the $ riangle A B C$ is

Question

Consider a branch of the hyperbola $x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0$ with vertex at the point $A$. Let $B$ be one of the end points of its latus rectum. If $C$ is the focus of the hyperbola nearest to the point $A$, then the area of the $	riangle A B C$ is

Aicharya · Accepted Answer

The given equation can be rewritten as $\frac{(x-\sqrt{2})^2}{1}-\frac{(y+\sqrt{2})^2}{2}=1$ for $A(x, y)$,

$ \begin{aligned} & e=\sqrt{1+\frac{2}{4}}=\sqrt{\frac{3}{2}} \\ & \therefore x-\sqrt{2}=2 \Rightarrow x=2+\sqrt{2} \\ & \text { For } C(x, y), x-\sqrt{2}=a e=\sqrt{6} \\ & \Rightarrow \quad x=\sqrt{6}+\sqrt{2} \\ & \text { Now, } A C=\sqrt{6}+\sqrt{2}-2-\sqrt{2}=\sqrt{6}-2 \\ & \qquad B C=\frac{b^2}{a}=\frac{2}{2}=1 \\ & \text { Area of } \Delta A B C=\frac{1}{2} \times(\sqrt{6}-2) \times B C \\ & =\frac{1}{2} \times(\sqrt{6}-2) \times 1=\sqrt{\frac{3}{2}}-1 \end{aligned} $

Consider a branch of the hyperbola $x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0$ with vertex at the point $A$.…

Solution