First slide

Hyperbola in conic sections

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 near $A,$ then area of the $∆ A B C$ is:

Moderate

A

$1 - \sqrt{\frac{2}{3}}$
B

$\sqrt{\frac{3}{2}} - 1$
C

$1 + \sqrt{\frac{2}{3}}$
D

$\sqrt{\frac{3}{2}} + 1 .$

Solution

Equation of the hyperbola is $\frac{(x - \sqrt{2})^{2}}{4} - \frac{(y + \sqrt{2})^{2}}{2} = 1$
Which can be written as
$\frac{X^{2}}{4} - \frac{Y^{2}}{2} = 1 where x = X + \sqrt{2}, y = Y - \sqrt{2} .$
Eccentricity of the hyperbola is $\sqrt{\frac{4 + 2}{4}} = \sqrt{\frac{3}{2}}$

Area of the $Δ A B C = \frac{1}{2} (A C) (B C)$
$\begin{array}{l} = \frac{1}{2} (a e - a) \times \frac{b^{2}}{a} where a^{2} = 4, b^{2} = 2 \\ = \frac{1}{2} [2 \sqrt{\frac{3}{2}} - 2] \times \frac{2}{2} = \sqrt{\frac{3}{2}} - 1 \end{array}$

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