Question

A hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is drawn along with its conjugate hyperbola. The foci points of both hyperbolas are connected as shown. Then $S_1{S}_3{S}_2{S}_4$ always forms a

• A. Rhombus
• B. Trapezium
• C. Square
• D. None of the above

Answer 1

The correct option is C

Square

Here the axes forms the diagnols of the quadrilateral $S_1{S}_3{S}_2{S}_4$. since the axes are perpendicular to each other we can say diagonals are $\perp$ or to each other. ............(1)

Length of the diagonals $S_1{S}_2=2b{e}_2$

=$2b.\sqrt{\frac{a^2+b^2}{b^2}}$

=2$\sqrt{a^2+b^2}$

Length of the diagonals $S_3{S}_4=2b{e}_1$

=2a$\sqrt{\frac{a^2+b^2}{a^2}}$

=2$\sqrt{a^2+b^2}$

i.e., Diagonals are having equal lengths ......(2)

Condition (1) and (2) imply that

$\square S_1{S}_3{S}_2{S}_4$ is a square.