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JEE Advanced 2008Paper 2
Exam
JEE Advanced
Year
2008
Paper
Paper 2
Subject
Mathematics
Chapter
Hyperbola

JEE Advanced 2008 · Paper 2 · Mathematics · Hyperbola

Consider a branch of the hyperbola x^2-2 y^2-2 2 x-4 2 y-6=0 with...

Consider a branch of the hyperbola x2−2y2−22x−42y−6=0x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0 with vertex at the point AA. Let BB be one of the end points of its latus rectum. If CC is the focus of the hyperbola nearest to the point AA, then the area of the △ABC\triangle A B C is

Options

  1. A1−231-\sqrt{\frac{2}{3}} sq unit
  2. B32−1\sqrt{\frac{3}{2}}-1 sq unit
  3. C1+231+\sqrt{\frac{2}{3}} sq unit
  4. D32+1\sqrt{\frac{3}{2}}+1 sq unit

Correct answer

B. 32−1\sqrt{\frac{3}{2}}-1 sq unit

Step-by-step solution

The given equation can be rewritten as (x−2)21−(y+2)22=1\frac{(x-\sqrt{2})^2}{1}-\frac{(y+\sqrt{2})^2}{2}=1 for A(x, y),,e=1+24=32∴x−2=2⇒x=2+2 For C(x,y),x−2=ae=6⇒x=6+2 Now, AC=6+2−2−2=6−2BC=b2a=22=1 Area of ΔABC=12×(6−2)×BC=12×(6−2)×1=32−1\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}
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