First slide

Hyperbola in conic sections

Question

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

Moderate

A

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

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

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

$\sqrt{3 / 2} + 1$

Solution

$\begin{array}{l} x^{2} - 2 y^{2} - 2 \sqrt{2} x - 4 \sqrt{2} y - 6 = 0 \\ or (x^{2} - 2 \sqrt{2} + 2) - 2 (y^{2} + 2 \sqrt{2} y + 2) = 4 \\ or \frac{(x - \sqrt{2})^{2}}{4} - \frac{(y + \sqrt{2})^{2}}{2} = 1 \end{array}$
$Now B is one of the end points of its latus rectum and C is the focus of the hyperbola nearest to the vertex A.$
$Clearly, area of LABC does not change if we consider similar$
$hyperbola with center at (0, 0) or hyperbola \frac{x^{2}}{4} - \frac{y^{2}}{2} = 1$
$Here vertex is A (2, 0) .$
$∴ a^{2} e^{2} = a^{2} + b^{2} = 6$
$So, one of the foci is B (\sqrt{6}, 0), point C is (a e, b^{2} / a) or (\sqrt{6}, 1) .$
$\begin{array}{l} ∴ A B = \sqrt{6} - 2 and B C = 1 \\ ∴ Area of Δ A B C = \frac{1}{2} A B \times B C = \frac{1}{2} (\sqrt{6} - 2) \times 1 \end{array}$
$= \sqrt{\frac{3}{2}} - 1 sq.units.$

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