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10-2. Parabola, Ellipse, Hyperbola
normal
Area of the quadrilateral formed with the foci of the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ and $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = - 1$ is
A
$4(a^2 + b^2)$
B
$2(a^2 + b^2)$
C
$(a^2 + b^2)$
D
$\frac{1}{2}$$(a^2 + b^2)$
Solution
Given hyperbolas are conjugate and the quadrilateral formed by their foci is a square
now $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ and $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = – 1$
$e_1^2 = 1 + \frac{{{b^2}}}{{{a^2}}}$ ; $e_2^2 = 1 + \frac{{{a^2}}}{{{b^2}}}$ ;
$e_1^2\,e_2^2 = \frac{{{{({a^2} + {b^2})}^2}}}{{{a^2}{b^2}}}$ ;
$e_1e_2 =\frac{{{a^2} + {b^2}}}{{ab}}$
$A = $ $\frac{{(2a{e_1})(2b{e_2})}}{2}$
$= 2abe_1e_2 $
$=\frac{{2ab({a^2} + {b^2})}}{{ab}}$
Standard 11
Mathematics
